Differential Equation Rally

Big Idea

The general solution to the differential equation

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + cy = 0$

is given by:

$y = C_1 e^{\alpha x} + C_2 e^{\beta x}$ if the roots $\alpha \neq \beta$ are real and distinct,
$y = (C_1x + C_2) e^{\alpha x}$ if the roots $\alpha = \beta$ are real and repeated,
$y = e^{px}(C_1 \cos qx + C_2 \sin qx)$ if the roots $p \pm iq$ are complex conjugates,

where $\alpha, \beta$ are the roots (i.e. the auxiliary roots) of the characteristic equation or auxiliary equation

$am^2 + bm + c = 0.$

In the first case, we usually choose $\alpha < \beta$ for purely aesthetic purposes. We will also revise previous topics in this post.

Questions

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{3x^2}{2y}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{3x^2}{2y} \\ \int 2y\, \mathrm dy &= \int 3x^2\, \mathrm dx \\ y^2 &= x^3 + C. \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - 3y = e^{3x} \cos(x).$

(Click for Solution)

Solution. We first identify $P(x) = -3$ and $Q(x) = e^{3x} \cos(x)$ . We compute the integrating factor

$I(x) = e^{\int -3\, \mathrm dx} = e^{-3x}.$

Thus, the general solution is given by

$\begin{aligned} e^{-3x} \cdot y &= \int e^{-3x} \cdot e^{3x}\cos(x)\, \mathrm dx \\ &= \int \cos(x)\, \mathrm dx \\ &= \sin(x) + C. \end{aligned}$

Question 3. Solve the differential equation

$\displaystyle 3\frac{\mathrm d^2 y}{\mathrm dx^2} + 2 \frac{\mathrm dy}{\mathrm dx} - y = 0.$

(Click for Solution)

Solution. The auxiliary equation $3m^2 + 2m - 1 = 0$ has auxiliary roots $m = -1, 1/3$ . Since the roots are real and distinct, the general solution is given by $y = C_1 e^{-x} + C_2 e^{\frac 13 x}$ .

Question 4. Solve the differential equation

$\displaystyle 4\frac{\mathrm d^2 y}{\mathrm dx^2} - 4 \frac{\mathrm dy}{\mathrm dx} + y = 0.$

(Click for Solution)

Solution. The auxiliary equation $4m^2 - 4m + 1 = 0$ has auxiliary roots $m = 1/2$ . Since the roots are real and repeated, the general solution is given by $y = (C_1x + C_2) e^{\frac 12 x}$ .

Question 5. Solve the differential equation

$\displaystyle 5\frac{\mathrm d^2 y}{\mathrm dx^2} - 4 \frac{\mathrm dy}{\mathrm dx} + y = 0.$

(Click for Solution)

Solution. The auxiliary equation $5m^2 - 4m + 1 = 0$ has auxiliary roots $m = \frac 25 \pm \frac 15 i$ . Since the roots are complex conjugates, the general solution is given by $y = e^{\frac 25 x} ( C_1 \cos(\frac 15 x) + C_2 \sin(\frac 15 x) )$ .

Question 6. Solve the differential equation

$\displaystyle xe^y \frac{\mathrm dy}{\mathrm dx} - 2e^y = x^2 \ln(x), \quad x \neq 0.$

(Click for Solution)

Solution. Make the substitution $u = e^y$ so that $\displaystyle \frac{\mathrm du}{\mathrm dx} = e^y \frac{\mathrm dy}{\mathrm dx}$ , and hence

$\begin{aligned} x\frac{\mathrm du}{\mathrm dx} -2u = x^2 \ln(x). \end{aligned}$

Dividing by $x$ ,

$\begin{aligned} \frac{\mathrm du}{\mathrm dx} -\frac 2x \cdot u = x \ln(x). \end{aligned}$

We then identify $P(x) = -2/x$ and $Q(x) = x \ln(x)$ . Computing the integrating factor,

$\begin{aligned} I(x) &= e^{\int -\frac 2x\, \mathrm dx} = e^{-2 \ln|x|} = \displaystyle \frac 1{x^2}. \end{aligned}$

Thus, the general solution is given by

$\begin{aligned} \frac 1{x^2} \cdot u &= \int \frac 1{x^2} \cdot x \ln(x)\, \mathrm dx \\ &= \int \frac 1x \cdot \ln(x)\, \mathrm dx. \end{aligned}$

To simplify the right-hand side, we integrate by substitution. Making the substitution $v = \ln(x)$ so that $\mathrm dv = \frac 1x\, \mathrm dx$ ,

$\begin{aligned} \int \frac 1x \cdot \ln(x)\, \mathrm dx &= \int v\, \mathrm dv = \frac{v^2}{2} + C = \frac{(\ln(x))^2}{2} + C. \end{aligned}$

Finally, recalling that $u = e^y$ ,

$\displaystyle \frac 1{x^2} \cdot e^y = \frac{(\ln(x))^2}{2} + C.$

Question 7. Modify your strategy in Question 6 to solve the differential equation

$\displaystyle xe^y \frac{\mathrm dy}{\mathrm dx} - 2e^y = x^n \ln(x), \quad x \neq 0,$

where $n \neq 2$ is any integer.

(Click for Solution)

Solution. We leave it as an exercise in integration by parts to verify that

$\displaystyle \frac 1{x^2} \cdot e^y = \int x^{n-3} \ln(x)\, \mathrm dx = \frac{ x^{n-2} \ln(x)}{n-2} - \frac{ x^{n-2}}{(n-2)^2} + C.$

—Joel Kindiak, 18 Apr 25, 1211H

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