Ordinary differential equations

Each numbered equation in this section represents a unique type of differential equation. For a thorough review of this section, be sure to know how to solve each type of differential equation.

1. Define a homogeneous function. Provide an example of a homogeneous functions. [answer]

   Simply put, \[f(sx_1, \dots, sx_n) = s^k f(x_1,\dots, x_n)\] is a homogeneous function, where \(k\) is the "degree of homogeneity." For example, \(f(x,y,z) = x^5 y^2 z^3\) is homogeneous of degree 10 because \(f(\alpha x,\alpha y,\alpha z) = (\alpha x)^5 (\alpha y)^2 (\alpha z)^3 = \alpha^{10} f(x,y,z)\).

2. Define a homogeneous polynomial and provide an example. [answer]

   Similar to a homogeneous function, a homogeneous polynomial is defined by \[P(sx_1, \dots, sx_n) = s^k P(x_1,\dots, x_n)\] where \(P\) is some multivariate polynomial. Simply put, a polynomial is homogeneous if its terms all have the same degree. An example of a homogeneous polynomial is \(x^2 + 2xy + y^2\).

3. What defines the homogeneity of a rational function? [answer]

   A rational function is homogeneous if it consists of a ratio of two homogeneous polynomials.

4. Define the linearity of a function \(f(x)\). Linear functions obey the _________ principle. [answer]

   The properties \(f(x+y) = f(x) + f(y)\) and \(f(ax) = af(x)\) for all \(a\) together define linearity. Linear functions obey the superposition principle.

5. Is a homogeneous function always linear? Is a linear function always homogeneous? [answer]

   A homogeneous function is not always linear. For example, \(f(x,y,z) = x^5 y^2 z^3\) is homogeneous but not linear. However, a linear function is always homogeneous, because homogeneity is one of the two properties of linear functions.

6. What defines a homogeneous first-order ordinary differential equation? Provide an example of a first-order ODE that is homogeneous, as well as an example of one that is inhomogeneous. [answer]

   A homogeneous ordinary differential equation is of the form \(\frac{dy}{dx} = f(x,y) = f\bigg(\frac{y}{x}\bigg)\). An example of a first-order ODE that is homogeneous is \(\frac{dy}{dx} = \frac{\sin (y/x)}{y/x} + (y/x)^2\), while an example of a first-order ODE that is inhomogeneous is \(\frac{dy}{dx} = y+x\).

7. Classify the differential equation \begin{equation}\label{ode1}\tag{1} \frac{dy}{dx} = a(x) y\,. \end{equation} Find the general solution to equation (\ref{ode1}). [answer]

   This is a linear, homogeneous, separable ordinary differential equation. The general solution is \[y = \pm C e^{\int a(x)dx}\,.\]

8. Classify and solve \(\frac{dy}{dx} = [x + \cos(x) ]y\,.\) for \(y\). [answer]

   This is a linear, homogeneous, separable ordinary differential equation. The general solution is \[y = \pm C e^{\int a(x)dx}\,.\]

9. Classify and solve \(\frac{dy}{dx} = (y-x)^2\). The solution can remain in an implicit form. Hint: is there a transformation that can make this ODE separable? [answer]

   As written, this is a nonlinear, homogeneous, inseparable ordinary differential equation. However, it can be made separable by making the substitution \(z = y-x\). Then, the ODE becomes \[\frac{dz}{dx} = z^2-1,\] and the general solution is \[\frac{\ln{(z-1)}}{\ln{(z+1)}} = 2x + C\,.\]

10. Classify and solve \(\frac{dy}{dt} = \cos(y-t)\). Follow the hint above. [answer]

    As written, this is a nonlinear, homogeneous, inseparable ordinary differential equation. However, it can be made separable by making the substitution \(z = y-t\). Then, the ODE becomes \[\frac{dz}{dx} = \cos(z)-1,\] and the general solution is found by performing the integral \[\int \frac{dz}{\cos z- 1} = t.\] I used Wolfram to calculate the integral (It is doable, but involves a trigonometric substitution). The solution is \[\cot{\frac{z}{2}} = t,\] or in an explicit form, \[z = 2 \text{arccot} (t),\]

11. Classify and solve \( y + \sqrt{xy} = x\frac{dy}{dx}\), where \(x>0\). Hint: is there a transformation that can make this ODE separable? This time it a multiplicative transformation, unlike that addititive transformations in the previous two examples. [answer]

    First the ODE is rearranged into the form, \begin{align*} x\frac{dy}{dx} &= y + \sqrt{xy}\\ \frac{dy}{dx} &= \frac{y}{x} + \frac{\sqrt{xy}}{x}\\ \frac{dy}{dx} &= \frac{y}{x} + {\sqrt{\frac{y}{x}}} \end{align*} From the last equation it can be seen that the ODE is homogeneous (because it is of the form \(dy/dx= f(y/x)\)). Making the substitution \(y/x = u\) results in \begin{align*} x\frac{du}{dx} + u &= u + \sqrt{u}\\ x\frac{du}{dx} &= \sqrt{u}\\ \int u^{-1/2} du &= \int \frac{dx}{x}\\ 2 u^{1/2} &= \ln{x} + C\\ y^{1/2}/x^{1/2} &= \frac{1}{2}(\ln{x} + C)\\ y &= x(\frac{1}{2}\ln{x} + C)^2 \end{align*}

12. Classify the differential equation \begin{equation}\label{ode2}\tag{2} \frac{dy}{dx} = a(x) y + b(x)\,. \end{equation} Find the general solution to equation (\ref{ode2}) by assuming the form of solution to equation (\ref{ode1}), but with \(C = C(x)\). What is the name of this solution? [answer]

    This is a linear, inhomogeneous first-order ordinary differential equation.

    The form of solution is assumed to be \(y = C(x)e^{\int a(x)dx}\). Inserting this solution into equation (\ref{ode2}) and applying the product rule gives \begin{align*} e^{\int a(x)dx}\frac{d}{dx} C(x) + C(x) \frac{d}{dx} e^{\int a(x)dx} &= a(x)y + b(x)\\ \end{align*} Note that \(\frac{d}{dx} e^{\int a(x)dx} = e^{\int a(x)dx} \frac{d}{dx}\int a(x)dx = a(x)e^{\int a(x)dx} \). Therefore, \begin{align*} e^{\int a(x)dx}\frac{d}{dx} C(x) + C(x)a(x)e^{\int a(x)dx} &= a(x)y + b(x)\\ &= a(x)C(x)e^{\int a(x)dx} +b(x) \end{align*} Canceling the common term above gives \begin{align*} e^{\int a(x)dx}\frac{d\, C(x)}{dx} &=b(x) \end{align*} Solving for \(C(x)\) gives \begin{align*} C(x) &= \int b(x) e^{-\int a(x)dx} dx \end{align*} Substituting this equation for \(C(x)\) into the assumed form of solution gives \begin{align*} y = e^{\int a(x) dx}\bigg[\int b(x) e^{-\int a(x)dx} dx\bigg] \end{align*} This is called the Cauchy equation.

13. Classify and solve \((1 + x^2)y' + 2xy = \cos x\). [answer]

    This is a linear, inhomogeneous first-order ordinary differential equation. Rewriting it as \(y' = \frac{\cos x}{1 + x^2} - \frac{2xy}{1 + x^2} \) renders it in the form of \(y' = a(x)y + b(x)\). Thus the Cauchy formula \begin{align*} y = e^{\int a(x) dx}\bigg[\int b(x) e^{-\int a(x)dx} dx\bigg] \end{align*} can be applied with \begin{align*} a(x)&= -\frac{2x}{1 + x^2}\\ b(x)&=\frac{\cos x}{1 + x^2}\,. \end{align*} Noting that \begin{align*} \int a(x) dx &= -\int \frac{2x}{1 + x^2} dx\\ &= -\ln(1+x^2) + C\quad\text{and}\\ \int b(x) e^{-\int a(x)dx}dx &= \int \frac{\cos(x)}{1 + x^2} e^{\ln(1+x^2)} dx\\ &=\int \cos{x} dx = \sin{x} + C \end{align*} Thus Cauchy equation is \begin{align*} y &= e^{\ln(1+x^2)^{-1}}(\sin x + C)\\ &=\frac{\sin x}{1+x^2} + C \end{align*} where \(C\) above varies form line to line (I know it is not technically correct, but one can absorb any arbitrary constant in \(C\)).

14. Classify the differential equation \begin{equation}\label{bern}\tag{3} \frac{dy}{dx} = a(x)y + b(x)y^n \end{equation} where \(n\neq 0,1\) is real. What is the name of this equation? What substitution does one make to facilitate its solution? [answer]

    This is a nonlinear, inhomogeneous first-order ordinary differential equation. It is called the Bernoulli equation.

    To solve this equation, one should seek a substitution that makes the equation linear. Let \(y=z^\alpha\), and therefore \(dy/dx = \alpha z^{\alpha-1}dz/dx\) Then equation (\ref{bern}) becomes \begin{align*} \alpha \frac{dz}{dx} = a(x)z + b(x)z^{\alpha n -\alpha + 1} \end{align*} In order for the above equation to be reduced to a linear inhomogeneous equation [see equation (\ref{ode2})], one should set the exponent of \(z\) multiplying \(b(x)\) equal to \(0\), which results in \begin{align*} \alpha = \frac{1}{1-n} \end{align*}

15. Classify and solve \(xy dy = (y^2 + x)dx\). [answer]

    Rearrange the equation into its standard form: \( \frac{dy}{dx} = \frac{y^2 + x}{xy} = \frac{y}{x} + \frac{1}{y}\). This is a nonlinear inhomogeneous ordinary differential equation, \(\frac{dy}{dx} = a(x)y + b(x)y^n\) with \(a(x) = \frac{1}{x}\), \(b(x) = 1\), and \(n = -1\). To transform the ODE into a linear one, set \(y=z^\alpha\), where \(\alpha = \frac{1}{1-n} = \frac{1}{2},\) resulting in \begin{align*} \alpha\frac{dz}{dx} &= a(x)z + b(x)\\ \frac{1}{2}\frac{dz}{dx} &= \frac{1}{x}z + 1\\ \frac{dz}{dx} &= \frac{2}{x}z + 2\\ \end{align*} \(\frac{dz}{dx} = \frac{2}{x}z + 2\) is now a linear inhomogeneous ordinary differential equation whose solution is given by the Cauchy equation: \begin{align*} z &= e^{\int \frac{2}{x}dx}\, \bigg( 2\int e^{-\int \frac{2}{x}dx}dx\bigg).\\ &=-2x + C \end{align*} The solution to for \(y\) is thus \(y= \pm\sqrt{C-2x}\).

    In my ODE notes from UT Dallas, I had \(y= \pm\sqrt{Cx^2-2x}\), but I disagree with the way the constants of integration were defined in the Cauchy equation that lead to this answer.

16. Classify the differential equation \begin{equation}\label{exact}\tag{4} M(x,y)dx + N(x,y)dy = 0\,,\quad \text{where } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\,. \end{equation} What is the general solution of equation (\ref{exact}) in integral form? [answer]

    This is an exact equation, and the solution in integral form is \(F = \int \frac{\partial F}{\partial x}dx = \int M dx\). Equivalently, the solution is \(F = \int \frac{\partial F}{\partial y}dy = \int N dy\). One should take the integral over whichever variable is easier to integrate over. After the taking the integral over one variable, the unknown constant is determined by taking the derivative of \(F\) with respect to the other variable, and setting this equal to \(M\) if the other variable is \(x\), or \(N\) if the other variable is \(y\).

17. Classify and solve \((4x^3 + 3y)dx + (3x + 4y^3)dy = 0\). [answer]

    This is an exact differential equation because \begin{align*} \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 3 \end{align*} Therefore the solution is \begin{align*} F(x,y) = \int \frac{\partial F}{\partial x} dx &= \int (4x^3 + 3y)dx\\ &= x^4 + 3xy + C(y) \end{align*} To determine \(C(y)\), take the partial derivative with respect to \(y\) and set this equal to \(N = 3x + 4y^3\): \begin{align*} 3x + C'(y) &= 3x + 4y^3\\ C'(y) &= 4y^3\\ C(y) &= y^4 \end{align*} Thus the solution is \begin{align*} F(x,y) = y^4 + x^4 + 3xy\,. \end{align*}

18. How does one solve \begin{equation}\label{integratingfactor}\tag{5} M(x,y)dx + N(x,y)dy = 0\,,\quad \text{but now where } \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\,? \end{equation} What is the name of this method? [answer]

    One can solve equation (\ref{integratingfactor}) by multiplying through by \(\mu(x)\) or \(\mu(y)\), and then forcing that the mixed partials of \(F\) be equal. Note that multiplying through by \(\mu(x,y)\) leads to a partial differential equation, which is not desired in this context.

    In this derivation, multiply through by \(\mu(x)\), \begin{align*} \mu(x)M(x,y)dx + \mu(x)N(x,y)dy = 0\,, \end{align*} and then set \begin{align} \frac{\partial}{\partial y}[\mu(x)M(x,y)] &= \frac{\partial}{\partial x}[\mu(x)N(x,y)] \notag\\ \mu \frac{\partial M}{\partial y} &= \frac{d\mu}{dx} N + \mu \frac{\partial N}{\partial x}\notag\\ \frac{d\mu}{dx} N&= \mu \bigg[\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \bigg] \notag\\ \frac{1}{\mu}\frac{d\mu}{dx} &= \frac{\partial M/\partial y - \partial N/\partial x}{N}\tag{i}\label{interatheraway} \end{align} Equation (\ref{interatheraway}) can be integrated and solved for \(\mu\). Thus \(\mu\) is called the integrating factor, and this method is called the ''integrating factor method.''

    Note that the right-hand side of equation (\ref{interatheraway}) must be only a function of \(x\), in the case above. If it is not (i.e., if it is a function of \(y\) as well), then try multiplying the original equation by \(\mu_y\), for which \begin{align*} \frac{1}{\mu}\frac{d\mu}{dy} &= \frac{\partial N/\partial x- \partial M/\partial y}{M} \end{align*}

19. Classify and solve \((x+2)\sin y\, dx + x \cos y \, dy=0\). [answer]

    This looks like it might be in the form of an exact equation, but since \(\partial M/\partial y = (x+2) \cos y \neq \partial N/\partial x = \cos y \), it requires the integrating factor method to turn it into an exact equation. An integrating factor \(\mu(x)\) is attempted. \begin{align*} \frac{1}{\mu}\frac{d\mu}{d x} &= \frac{(x+2) \cos y - \cos y}{x \cos y} = 1 + \frac{1}{x}\\ \int \frac{d\mu}{\mu} &= \int \bigg(1 + \frac{1}{x}\bigg)dx\\ \ln \mu &= x + \ln x\\ \mu &= e^x e^{\ln x}= xe^x \end{align*} Now the original ODE becomes \begin{align*} xe^x(x+2)\sin y\, dx + x^2 e^x \cos y \, dy=0 \end{align*} The solution is \begin{align*} F(x,y) = \int \frac{\partial F}{\partial x}dx = \dots \end{align*} However, before proceeding with this option, note that in this case that it is much easier to find \begin{align*} F(x,y) = \int \frac{\partial F}{\partial y} dy = \int x^2 e^x \cos y\, dy = x^2 e^x \sin y + C(x)\,. \end{align*} The constant \(C(x)\) is found by differentiating \(F\) with respect to \(x\) and setting this equal to \(x e^x (x+2)\sin y\): \begin{align*} \frac{\partial F}{\partial x} = (2x e^x + x^2 e^x) \sin y + C'(x) &= x e^x (x+2)\sin y \end{align*} Thus \[C'(x) = x^2 e^x \sin y + 2x e^x \sin y - (2x e^x + x^2 e^x) \sin y = 0,\] and thus \(C(x) = \text{const}\). The solution is therefore \(F(x,y) = x^2 e^x \sin y + \text{const}\).

20. Classify the differential equation \begin{equation}\label{secondorder}\tag{6} ay'' + b' + cy = 0\,, \end{equation} where \(a\), \(b\), and \(c\) are constants, and where the prime \('\) will now be used for notational ease to signify derivative. How does one solve this equation? There are the three possible cases that emerge. What are they? [answer]

    This is a linear, second-order ordinary differential equation with constant coefficients. It is solved by making the substitution \(e^{rx}\), which leads to a quadratic equation in \(r\). Note that this generalizes to higher derivatives, leading to polynomials of higher degree to solve.

    Three cases emerge.

    1. The solutions \(r\) are real (discrimant \(b^2-4ac>0\). In this case, the solution to the 2nd order ODE is exponential decay and/or growth.
    2. The solutions \(r\) are complex (discrimant \(b^2-4ac<0\). In this case, the solution to the 2nd order ODE is waves, which can either be written as complex exponentials, or as sines and cosines.
    3. The solutions \(r\) are equal, i.e., a double root (discrimant \(b^2-4ac=0\). In this case, the second solution gets multiplied by \(x\times\) the first. This will be proved later.

21. Classify and solve \(y''' + 4y'' + 9y' + 10y = 0\). [answer]

    This is a linear, third-order ordinary differential equation. Follow the procedure above and obtain the characteristic cubic equation in \(r\), \(r^3 + 4r^2 + 9r + 10 = 0\). The three solutions of this equation are \( r= -2\), \( r= -1 + 2i\), and \( r= -1 -2i\). Thus the solution is \(y= C_1 e^{-2x}+ C_2 e^{-x}\cos(2x)+ C_3 e^{-x}\sin(2x) \).

22. Classify and solve \(y''' + 6y'' + 12y' + 8y = 0\). [answer]

    This is also a a linear, third-order ordinary differential equation. Following the same procedure as above, obtain the characteristic cubic equation in \(r\), \(r^3 + 6r^2 + 12r + 8 = 0\). The three solutions of this equation are the triple root \( r= -2\). Thus the solution is therefore \(y= C_1 e^{-2x}+ C_2 xe^{-2x} + C_3 x^2e^{-2x}\).

23. What is the definition of the Wronskian? What information does it provide? [answer]

    The Wronskian \(W\) is defined as \begin{align*} W = \begin{vmatrix} y_1(x) & y_2(x)\\ y_1'(x) & y_2'(x) \end{vmatrix} =y_1 y_2' - y_2 y_1'\,, \end{align*} where \(y_1\) and \(y_2\) are two solutions to a second-order linear ordinary differential equation, and where \('\) signifies derivative. Two solutions of the same equation are called a fundamental pair if \(W \neq 0\).

    There is another way to calculate \(W\): Abel's formula for the Wronskian: \(W = \int e^{-\int p(x) dx}\), where the ODE is written as \(y'' + p(x)y' +q(x)y = 0\). However, I don't think it's worth reviewing.

24. Determine whether \(y_1 = x\) and \(y_2 = \ln x\) are a fundamental pair, and if so, on what interval. What about \( y_1 = \arccos \frac{x}{\pi}\) and \(y_2 = \arcsin \frac{x}{\pi}\)? [answer]

    The Wronskian for the first case is \(W= \ln x \), which is never 0. Thus \(y_1 = x\) and \(y_2 = \ln x\) are a fundamental pair for all \(x>0\) (since \(\ln x\) is defined on that interval).

    The Wronskian for the second case is \(W = \frac{\arccos(x/\pi)}{\pi\sqrt{1-x^2}} + \frac{\arcsin(x/\pi)}{\pi\sqrt{1-x^2}} \), which (I don't think) has a solution. Thus the solutions are a fundamental pair for \(x\in (-\pi,\pi)\).

25. ☸ Classify the differential equation \begin{equation}\label{homo2}\tag{7} a(x) y'' + b(x) y' + c(x) y = 0\,. \end{equation} Given one solution \(y_1\) to equation (\ref{homo2}), how can the other solution \(y_2\) be found? What is this method called? Hint: Let \(y_2 = uy_1\). Insert this into equation (\ref{homo2}). Introduce \(z= u'\) and then let \(z= u'\) (This is where the name of the method comes from). Solve for \(z\), integrate, and find \(u\). \(y_2\) is then found because \(y_2 = uy_1\). I doubt this would be on the exam as it is too involved. [answer]

    Assume that \(y_2 = u y_1\). Then the derivatives of \(y_2\) are \begin{align*} y_2' &= u' y_1 + u y_1'\\ y_2'' &= u'' y_1 +u' y_1' + u' y_1' + u y_1''\,. \end{align*} Inserting these relations into equation (\ref{homo2}) results in \begin{align*} a(x)(u'' y_1 +u' y_1' + u' y_1' + u y_1'') + b(x) (u' y_1 + u y_1') + c(x) uy_1 &= 0\,. \end{align*} This equation is regrouped: \begin{align*} u[a(x) y_1'' + b(x) y_1' + c(x)] + a(x)u'' y_1 + 2a(x)u' y_1' + b(x)u' y_1 &= 0\,. \end{align*} The first term in \([...]\) is \(0\) by equation (\ref{homo2}). Therefore, \begin{align*} a(x)u'' y_1 + u'[2a(x) y_1' + b(x)y_1 ] &= 0\,. \end{align*} Now, introduce the parameter \(z = u'\), which kicks all the derivatives above down one: \begin{align*} a(x) y_1 \frac{dz}{dx} + z[2a(x) y_1' + b(x)y_1 ] &= 0\,.\\ \frac{dz}{z} &= \bigg[-\frac{2 y_1'}{y_1} - \frac{b(x)}{a(x)} \bigg]dx \end{align*} Integrating results in \begin{align*} \ln z &= -2 \int \frac{ y_1'}{y_1}dx - \int \frac{b(x)}{a(x)}dx \\ &= -2 \int \frac{dy_1}{dx}\frac{1}{y_1}dx - \int \frac{b(x)}{a(x)}dx \\ &= -2 \int\frac{dy_1}{y_1} - \int \frac{b(x)}{a(x)}dx \\ &= \ln (y_1^{-2}) - \int \frac{b(x)}{a(x)}dx \,. \end{align*} Exponentiating gives \begin{align*} z &= y_1^{-2}e^{\int \frac{b(x)}{a(x)} dx}\,.\\ u &= \int z dx = \int y_1^{-2}e^{\int \frac{b(x)}{a(x)} dx} dx\,.\\ y_2 &= y_1u = y_1\int y_1^{-2}e^{\int \frac{b(x)}{a(x)} dx} dx\,. \end{align*}

    This method is called reduction of order.

26. Given that \(y_1 = x\) is a solution to \(x^2 y'' -x(x+2) y' + (x+2) y = 0\), classify this equation and find the general solution. [answer]

    This is a linear homogeneous second-order ordinary differential equation with nonconstant coefficients. Reduction of order should be used: \begin{align*} y_2 &= y_1\int y_1^{-2}e^{\int \frac{b(x)}{a(x)} dx} dx\\ &= x \int x^{-2}e^{-\int \frac{x(x+2)}{x^2} dx} dx\\ &= x \int x^{-2}e^{-\int (1 + 2/x)dx} dx\\ &= x \int x^{-2}e^{-x + \ln (x^{-2})} dx\\ &= x \int e^{-x} dx = xe^{-x}\\ \end{align*}

27. Classify the differential equation \begin{equation}\label{inhomo2} \tag{8} ay'' + by' + cy = g(x)\,. \end{equation} List the two ways to solve this equation. [answer]

    This is linear inhomogeneous second-order ordinary differential equation with constant coefficients. The two methods to solve this are (1) the method of undetermined coefficients and (2) the variation of parameters:

    First let us discuss the method of unetermined coefficients, which works when the right-hand side \(g(x)\) is exponential, sinusoidal, polynomial, or a product or sum of exponentials, sinusoidals, and polynomials.

    1. If \(g(x) = e^{kx}p(x)\), where \(p\) is a polynomial of degree \(n\), then use as the form of particular solution \(y_p(x) = e^{kx}q(x)\), where \(q\) is the \(n\)th degree polynomial. The coefficients of \(q\) are then determined by substituting it into equation (\ref{inhomo2}).
    2. If \(g(x) = e^{kx}p(x)\cos(mx)\), or if \(g(x) = e^{kx}p(x)\sin(mx)\) where \(p\) is a polynomial of degree \(n\), then use as the form of particular solution \(y_p(x) = e^{kx}q(x)\cos (mx) + e^{kx}s(x)\sin (mx) + \), where \(q\) and \(s\) are \(n\)th degree polynomials with different coefficients.

    Importantly, in either case above, if any term of the particular solution is also a solution of the homogeneous equation, then the form of solution should be multiplied by \(x\) (if the particular solution coincides with the homogeneous solution once), or \(x^2\) (if the particular solution coincides with the homogeneous solution twice). This is called resonance.

28. Classify and solve \(y'' + 4y = e^{3x}\). [answer]

    \(y = y_h + y_p = C_1\cos 2x + C_2 \sin 2x + \frac{1}{13}e^{3x}\).

29. Classify and solve \(y'' + y = \sin x\). [answer]

    See here for a video solution. There a resonance here of multiplicity \(1\). The solution is \(y = C_1 \cos x + C_2 \sin x - \frac{x}{2}\cos x\).

30. Classify and solve \(y'' - 4y' + 3y = 2xe^{x}\). [answer]

    The method of undetermined coefficients is used. However, one needs to be careful in this case, because the particular solution resonates with the homogeneous solution. Thus the particular solution be chosen to be \(y_p = xe^{-x}(Ax +B)\). See here for the complete solution.

31. Classify and solve \(y'' + 2y' + y = e^{-x}\). [answer]

    The method of undetermined coefficients is used. Again, one needs to be careful because the particular solution resonates with the homogeneous solution, this time with multiplicity 2. Thus the particular solution be chosen to be \(y_p = Ax^2e^{-x}\). See here for the complete solution.

32. Determine the form of trial solution for \(y'' -4y' + 13y = e^{2x}\cos 3x\). [answer]

    Note that the homogeneous solution has characteristic roots \(r = 2+3i\) and \(r = 2-3i\). The first of these roots coincides with the driving function, \(e^{2x}\cos 3x\). Thus the form of solution should be the non-resonant guess multiplied by \(x\), i.e., \(xe^{2x}(A\cos 3x + B\sin 3x)\).

33. What is the variation of parameters? Do not provide the full derivation, but provide the big picture (like, why is it called "variation of parameters"?). In what situations should the variation of parameters be used? [answer]

    Variation of parameters provides a more general way of solving \(y'' + p(x)y' + q(x)y= 0\), a linear 2nd order inhomogeneous ordinary differential eqution with non-constant coefficients. It can of course also be used to solve a linear 2nd order inhomogeneous ordinary differential eqution with constant coefficients, and in fact this is the method that must be used when the right-hand side \(g\) is more complicated than a sum or product of polynomials, trigonometric functions, or exponentials.

    To derive the variation of parameters, first one solves the homogeneous equation to obtain solutions \(y_1\) and \(y_2\). Then, one looks for the form of solution \(y_p= C_1(x)y_1(x) + C_2(x)y_2(x),\) which gives the method its name \(C_1\) and \(C_2\) are allowed to vary as functions, not constants. An additional constraint \begin{equation}\label{constrainterers}\tag{i} C_1'y_1 + C_2' y_2 = 0 \end{equation} is imposed. Taking the derivatives of \(y\), substituting the result into the ODE, and simplifying results in \begin{equation}\label{solutitors}\tag{ii} y_p= C_1'y_1' + C_2'y_2'\,. \end{equation} Combining equations (\ref{constrainterers}) and (\ref{solutitors}) results (by Cramer's rule) in a solution for \(y\): \begin{equation}\label{variationofparam}\tag{iii} y_p = -y_1 \int \frac{y_2 g}{W}dx + y_2 \int \frac{y_1 g}{W}dx\,. \end{equation} Just memorize equation (\ref{variationofparam}).

34. Solve \(y'' + 4y = \frac{3}{\sin x}\). [answer]

    See here.

35. Solve \(y'' - 2y' + y = \frac{e^t}{t^2 + 1}\). [answer]

    Using the variation of parameters, the particular solution is found by taking the integrals \begin{align*} y_p &= -y_1 \int \frac{y_2 g}{W}dx + y_2 \int \frac{y_1 g}{W}dx\\ &\dots\\ &= C_1 e^t + C_2 t e^t -\frac{e^t}{2}\ln (t^2 +1) + te^t \arctan t\,. \end{align*}

36. ☸ What is \begin{equation}\label{Cauchy-Euler}\tag{9} ax^2 y'' + bxy' + cy = 0 \end{equation} called? (Why is it a bad name)? What clever substitution does one make to go about solving it? [answer]

    Equation (\ref{Cauchy-Euler}) is known as Euler's equation. A better name is the "Cauchy-Euler equation,'' because Euler already has so many equations named after him (like \(\frac{\partial f}{\partial y}-\frac{d}{dx} \frac{\partial f}{\partial y'}=0\) from the calculus of variations and \(e^{i\pi}+1=0\), the identity containing the most important numbers of mathematics)!

    To solve equation (\ref{Cauchy-Euler}), one lets \(x\) be an exponential function of a new independent variable \(t\). That is, \(x=e^t\). Then the derivatives of \(y\) with respect to \(t\) are \begin{align*} \frac{dy}{dt} &= \frac{dy}{dx}\frac{dx}{dt} = \frac{dy}{dx} e^t = \frac{dy}{dx} x \\ \frac{d^2y}{dt^2} &= \frac{d}{dt} \frac{dy}{dt} = \frac{d}{dt} \bigg(\frac{dy}{dx} x\bigg) = x\frac{d^2y}{dtdx}+ \frac{dx}{dt}\frac{dy}{dx} \end{align*} Note that \(\frac{d^2y}{dtdx} = \frac{d^2y}{dxdt} = \frac{d}{dx}\big(\frac{dy}{dx}\frac{dx}{dt}\big) = \frac{d}{dx}\big(\frac{dy}{dx}\big)\frac{dx}{dt} =\frac{d}{dx}\big(\frac{dy}{dx}\big)x\). Therefore, \begin{align*} \frac{d^2y}{dt^2}&= x^2 \frac{d^2y}{dx^2} + x\frac{dy}{dx}\,. \end{align*} Solving the above relations for the derivatives with respect to \(x\) gives \begin{align*} \frac{dy}{dx} &= \frac{1}{x}\frac{dy}{dt}\\ \frac{d^2 y}{dx^2} &= \frac{1}{x^2}\bigg(\frac{d^2y}{dt^2} - x\frac{dy}{dx}\bigg) \end{align*} Thus equation (\ref{Cauchy-Euler}) is \begin{align*} ax^2 \frac{d^2y}{dx^2} + bx \frac{dy}{dx} + cy &=0\\ a\bigg(\frac{d^2y}{dt^2} - x\frac{dy}{dx}\bigg) + b\frac{dy}{dt} + cy &=0\\ a\bigg(\frac{d^2y}{dt^2} - \frac{dy}{dt}\bigg) + b\frac{dy}{dt} + cy &=0\\ a\frac{d^2y}{dt^2} + (b-a)\frac{dy}{dt} + cy &=0\\ \end{align*} Therefore, the introduction of \(x=e^t\) successfully converts equation (\ref{Cauchy-Euler}) into a constant-coefficient differential equation, where the independent variable is \(t\) instead of \(x\). Thus, the solution to \(a\frac{d^2y}{dt^2} + (b-a)\frac{dy}{dt} + cy =0\) is \(y(t)\) (and it is solved the same way as equation (\ref{secondorder}), only in \(t\), i.e., letting \(y = e^{rt}\)), which must be converted back to \(y(x)\) by letting \(t = \ln x\).

37. What are the three cases that arise when solving equation (\ref{Cauchy-Euler})? [answer]

    It was shown in the previous problem that Euler's equation is a constant-coefficient differential equation where \(r\) is the independent variable: \[a^2y'' + (b-a)y' + cy = 0\,.\] Three cases arise:

    1. \(a^2y'' + (b-a)y' + cy = 0\) has real roots \(r_1\) and \(r_2\), and the solution to Euler's differential equation is \begin{align*} y(t) &= C_1 e^{r_1t} + C_2 e^{r_2t}\\ \implies y(x)&= C_1 x^{r_1} + C_2 x^{r_2}\,. \end{align*}
    2. \(a^2y'' + (b-a)y' + cy = 0\) has a double roots \(r_1 = r_2 \equiv r\), and the solution to Euler's differential equation is \begin{align*} y(t) &= C_1 e^{rt} + C_2 t e^{rt}\\ \implies y(x)&= C_1 x^{r} + C_2 \ln{x} x^{r}\,. \end{align*}
    3. \(a^2y'' + (b-a)y' + cy = 0\) has a complex roots \(r_1 = \alpha + i\beta\) and \( r_2 = \alpha - i\beta \), and the solution to Euler's differential equation is \begin{align*} y(t) &= C_1 e^{\alpha t}\cos \beta t + C_2 e^{\alpha t} \sin {\beta t}\\ \implies y(x) &= C_1 \alpha t\cos (\beta \ln x) + C_2 \alpha t \sin ({\beta \ln t})\\ \end{align*}

    Thus, to remember how to solve equation (\ref{Cauchy-Euler}), one must simply remember that the substitution \(x =e^t\) converts equation (\ref{Cauchy-Euler}) into \(a^2y'' + (b-a)y' + cy = 0\).
38. Classify and solve \(2x^2 y'' + 3x y' -y = 0\). [answer]

    This is a Cauchy-Euler differential equation. Recalling that the substitution \(x =e^t\) converts equation (\ref{Cauchy-Euler}) into \(a^2y'' + (b-a)y' + cy = 0\), the coefficients \(a=2\), \(b = 3\) and \(c=-1\) are identified, and the constant-coefficient equation is \[2y'' + y' - y = 0,\] the solution of which is \begin{align*} y(t) = C_1 e^{t/2} + C_2 e^{-t} \end{align*} Making the substitution \(x = e^t\) gives the solution of the Cauchy-Euler differential equation: \begin{align*} y(x) = C_1 x^{1/2} + C_2 x^{-1}\,. \end{align*}

39. Let \(a_n\) be the expansion coefficients in a series solution \(y = \sum_n a_n x^n\). Define the radius of convergence of this sum. [answer]

    \begin{align*} R = \lim_{n\to\infty}\bigg\lvert\frac{a_n}{a_{n+1}}\bigg\rvert \end{align*}

40. Calculate the radius of convergence for \(e^x\). [answer]

    Noting that \(e^x = \sum_{n=0}\frac{x^n}{n!}\), the radius of convergence is \begin{align*} R &= \lim_{n\to\infty}\bigg\lvert\frac{(n+1)!}{n!}\bigg\rvert\\ &= \lim_{n\to\infty} (n+1) =\infty\,. \end{align*} That is, the series converges everywhere.

41. Calculate the radius of convergence for \(\frac{1}{1-x}\). [answer]

    Noting that \(\frac{1}{1-x} = \sum_{n=0}x^n \), the radius of convergence is \begin{align*} R &= \lim_{n\to\infty}\bigg\lvert\frac{(n+1)!}{n!}\bigg\rvert\\ &= \lim_{n\to\infty} 1 =1\,. \end{align*} That is, this series converges on the interval \(x =(-1 ,1)\).

42. Solve \(y'' + y = 0\) by series. [answer]

    Let \begin{align*} y &= \sum_{n=0}^{\infty} a_n x^n\\ y' &= \sum_{n=0}^{\infty}n a_n x^{n-1} = \sum_{n=1}^{\infty}n a_n x^{n-1} \\ y'' &= \sum_{n=0}^{\infty}n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty}n(n-1) a_n x^{n-2}\\ \end{align*} where it has been noted that the first term in the sum for \(y'\) is \(0\), and the first two terms in the sum for \(y''\) are \(0\). Substituting these sums into the ODE gives \begin{align*} \sum_{n=2}^{\infty}n(n-1) a_n x^{n-2} + \sum_{n=0}^{\infty} a_n x^n = 0 \end{align*} Now the indices of the first summation term shifted from \(n+2 \mapsto n\), which means that \(2\) must be added to each \(n\) in the summand to compensate: \begin{align*} \sum_{n=0}^{\infty}(n+2)(n+1) a_{n+2} x^{n} + \sum_{n=0}^{\infty} a_n x^n = 0 \end{align*} The two summations may now be combined under the same sum: \begin{align*} \sum_{n=0}^{\infty}[(n+2)(n+1) a_{n+2} + a_n ]x^{n} = 0 \end{align*} Since this relation holds for every \(n\), one can identify a recurrence relation: \begin{align*} (n+2)(n+1) a_{n+2} + a_n &= 0\\ a_{n+2} &= -\frac{a_n}{(n+2)(n+1)} \end{align*} Now for the fun part, let \(a_0 = 1\) and \(a_1 = 0\). Then all \(a_n\) for \(n\) odd vanishes, and \begin{align*} n&= 0:\quad a_{2} = -\frac{a_0}{(0 + 2)(0+1)}= -\frac{1}{1\cdot 2}\\ n&= 2:\quad a_{4} = -\frac{a_2}{(2+2)(2+1)} = \frac{1}{1\cdot 2 \cdot 3\cdot 4}\\ n&= 4:\quad a_{6} = -\frac{a_4}{(4+2)(4+1)} = -\frac{1}{1\cdot 2 \cdot 3\cdot 4 \cdot 5 \cdot 6}\\ \end{align*} Defining \(k= 2n\), the pattern is explicitly given by \(a_{2k} = (-1)^k/(2k)!\) and \(a_{2k+1} = 0\). The solution can be written as \[\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} x^{2k}\]

    Meanwhile, letting \(a_0 = 0\) and \(a_1 = 1\) all \(a_n\) for \(n\) even vanishes, and \begin{align*} n&= 1:\quad a_{3} = -\frac{a_1}{(1 + 2)(1+1)}= -\frac{1}{2\cdot 3}\\ n&= 3:\quad a_{5} = -\frac{a_3}{(3+2)(3+1)} = \frac{1}{1\cdot 2 \cdot 4\cdot 5}\\ n&= 5:\quad a_{7} = -\frac{a_5}{(5+2)(5+1)} = -\frac{1}{1\cdot 2 \cdot 3\cdot 4 \cdot 5 \cdot 6\cdot 7}\\ \end{align*} Thus the pattern is explicitly given by \(a_{2k+1} = (-1)^k/(2k+1)!\) and \(a_{2k} = 0\). The solution can be written as \[\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k+1}\]

    The general solution is therefore \[y = a_0 \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k}+ a_1\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k+1}\] which we recognize as linear combinations of sines and cosines.

43. What is the name of the ordinary differential equation \(y'' - xy = 0\)? What are some of its applications? Solve it by series. [answer]

    This is the Airy equation, which was developed by Airy to describe caustics in optics, like in a rainbow. It is also used in quantum mechanics, at the turning point in the WKB approximation.

    As before, let \begin{align*} y &= \sum_{n=0}^{\infty} a_n x^n\\ y' &= \sum_{n=0}^{\infty}n a_n x^{n-1} = \sum_{n=1}^{\infty}n a_n x^{n-1} \\ y'' &= \sum_{n=0}^{\infty}n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty}n(n-1) a_n x^{n-2}\\ \end{align*} where again it has been noted that the first term in the sum for \(y'\) is \(0\), and the first two terms in the sum for \(y''\) are \(0\). Substituting these sums into the Airy equation gives \begin{align*} \sum_{n=2}^{\infty}n(n-1) a_n x^{n-2} - x\sum_{n=0}^{\infty} a_n x^{n} &= 0\\ \sum_{n=0}^{\infty}(n+2)(n+1) a_{n+2} x^{n} - \sum_{n=0}^{\infty} a_n x^{n+1} &= 0 \end{align*} Now the indices must be shifted so the powers of \(x\) match in both sums. This is done by shifting the index of the sum on the right from \(n=0\mapsto 1\), and shifting the index of the sum on the left from \(n=0\mapsto 1\): \begin{align*} \sum_{n=0}^{\infty}(n+2)(n+1) a_{n+2} x^{n} - \sum_{n=1}^{\infty} a_{n-1} x^{n} = 0 \end{align*} In order to combine the sums, the lower index must also match. This is done by explicitly removing the \(n=0\) terms of the sum on the left: \begin{align*} 2 a_{2} + \sum_{n=1}^{\infty}(n+2)(n+1) a_{n+2} x^{n} - \sum_{n=1}^{\infty} a_{n-1} x^{n} = 0 \end{align*} The two summations may now be combined under the same sum: \begin{align*} 2 a_{2} + \sum_{n=1}^{\infty}[(n+2)(n+1) a_{n+2} -a_{n-1}]x^{n} = 0 \end{align*} Since this relation holds for every \(n\), \(a_2 = 0\), and \begin{align*} (n+2)(n+1) a_{n+2} -a_{n-1} &= 0\\ a_{n+2} &= \frac{a_{n-1}}{(n+2)(n+1)} \end{align*} First, let \(a_0 = 1\) and \(a_1 = 0\). Then all \(a_n\) for \(n\) odd vanishes, and \begin{alignat*}{2} n&= 1:\quad a_{3} = \frac{a_0}{(1 + 2)(1+1)}= \frac{1}{2\cdot 3} &&\qquad k=1\\ n&= 2:\quad a_{4} = \frac{a_1}{(2+2)(2+1)} = 0 &&\\ n&= 3:\quad a_{5} = \frac{a_2}{(3+2)(3+1)} = 0 &&\\ n&= 4:\quad a_{6} = \frac{a_3}{(4+2)(4+1)} = \frac{1}{2\cdot 3\cdot 5\cdot 6} &&\qquad k=2 \end{alignat*} The pattern is explicitly given by \(a_{k} = [2\cdot 3 \cdot 5\cdot 6 \cdot \dots \cdot (3k+1)3k]^{-1}\) . The solution can be written as \[y_1 = 1 + \sum_{k=1}^\infty \frac{1}{2\cdot 3 \cdot 5\cdot 6 \cdot \dots \cdot (3k+1)3k} x^{3k},\] where the term \(1\) appears because \(a_0 = 1\) is not included in the summation.

    Meanwhile, letting \(a_0 = 0\) and \(a_1 = 1\) gives \begin{alignat*}{2} n&= 1:\quad a_{3} = 0 &&\\ n&= 2:\quad a_{4} = \frac{1}{3\cdot 4} && \qquad k = 1\\ n&= 3:\quad a_{5} = 0 &&\\ n&= 4:\quad a_{5} = 0 && \\ n&= 5:\quad a_{4} = \frac{1}{3\cdot 4 \cdot 6\cdot 7} && \qquad k=2 \end{alignat*} The pattern is explicitly given by \(a_{3k-1} = [3\cdot 4 \cdot 6\cdot 7\cdot \dots \cdot 3k(3k+1)]^{-1}\). The solution can be written as \[y_2 = 1 + \sum_{k=1}^\infty \frac{1}{3\cdot 4 \cdot 6\cdot 7\cdot \dots \cdot } x^{3k+1},\] where again the term \(1\) appears because \(a_1 = 1\) is not included in the summation.

    The sum \(y_1 +y_2\) is the general solution to Airy's equation.

44. Solve \(y'' + xy' +y = 0\) by series. [answer]

    See here for the solution.

45. When solving equation (\ref{homo2}) (the linear, homogeneous second-order ordinary differential equation with non-constant coefficients \(a(x) y'' + b(x) y' + c(x)y = 0\)), what must one be weary of if there is an \(x_0\) such that \(a(x_0) = 0\)? What is the condition on \(x_0\) for equation (\ref{homo2}) to be solved by series? [answer]

    \(x_0\) is called a singular point. Specifically, if the following limits are finite, \(x_0\) is called a "regular singular point," and the method of Frobenius can be used to solve eqaution (\ref{homo2}). \begin{align*} \lim_{x\to x_0} & \frac{b(x)}{a(x)} (x-x_0)\\ \lim_{x\to x_0} &\frac{c(x)}{a(x)} (x-x_0)^2 \end{align*} If either of the limits above are not finite, \(x_0\) is called a "singular singular point" (very creative!).

46. Identify the differential equation \((1-x^2)y'' - 2xy' + \alpha(\alpha +1)y = 0\), and identify the singular point(s). Classify them as "regular singular" or "singular singular." [answer]

    The singular points are \(x_0 = 1\) and \(x_0 =-1\). For \(x_0 = 1\), \begin{align*} \lim_{x\to 1} \frac{-2x}{1-x^2}(x-1) &=\lim_{x\to 1} \frac{2x}{x+1} = 1 \\ \lim_{x\to 1} \frac{\alpha(\alpha +1)}{1-x^2}(x-1)^2 &=\alpha(\alpha +1)\lim_{x\to 1} \frac{(x-1)(x-1)}{(1-x)(1+x)} = - \alpha(\alpha +1)\lim_{x\to 1} \frac{(x-1)}{(1+x)} = 0 \end{align*} Therefore the point \(x=1\) is a regular singular point. Meanwhile, for \(x_0 =-1\), \begin{align*} \lim_{x\to -1} \frac{-2x}{1-x^2}(x+1) &=\lim_{x\to -1} \frac{2x}{x-1} = \frac{-2}{-2} = 1\\ \lim_{x\to -1} \frac{\alpha(\alpha +1)}{1-x^2}(x+1)^2 &=\alpha(\alpha +1)\lim_{x\to -1} \frac{(x+1)(x+1)}{(1-x)(1+x)} = - \alpha(\alpha +1)\lim_{x\to -1} \frac{(x+1)}{(1-x)} = 0 \end{align*} Therefore the point \(x=-1\) is also a regular singular point.

47. Solve \(2x^2 y'' + xy' - (1+x)y = 0\). [answer]

    See here for the solution.

48. Identify and solve \(x^2 y'' + xy' + (x^2-\nu^2)y = 0\) for \(\nu=0\). What is the name of the solution? Where does it appear in acoustics? [answer]

    This is Bessel's equation, and the solutions for \(\nu=0\) are the Bessel functions \(J_0\) and \(N_0\). This appears in acoustics as the radial eigenfunction of the Helmholtz equation in cylindrical coordinates for axisymmetric radiation. See here for the solution for \(J_0\).