Exercises: Linear and Separable ODEs

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Exercises related to solving linear and separable ODEs.

Remember: several of these exercises involve logarithms. Use absolute value signs inside the logarithm when they’re needed.

Find the general solution of the ODE below. $y' = y - 2$ $\text {General Solution: } \ \ y = C \answer {e^x} + \answer {2}$ (For definiteness, the expression you enter in the second blank should equal $2$ when $x = 0$ .)

This ODE is both linear and separable, so either approach will work.

Find the general solution of the ODE below. $x^2 y' + xy = 1$ $\text {General Solution: } \ \ y = \answer {\frac {\ln |x|}{x}} + \frac {C}{x}$ (For definiteness, the function you enter for your answer should equal $0$ at $x=1$ . If that’s not the case, you might need to rewrite your solution and redefine the constant $C$ .)

This is a linear ODE.

Find the general solution of the ODE below. $y y' = 4x$ $\text {General Solution: } \ \ y^2 = \answer {4 x^2} + C$ (For definiteness, the function you enter for your answer should equal $0$ at $x=0$ . If that’s not the case, you might need to rewrite your solution and redefine the constant $C$ .)

This is a separable ODE. In the form written, it’s already separated.

$y'-3y=xe^{2x}$ $\text {General Solution: } \ \ y = C\answer {e^{3x}}-(x+1)\answer {e^{2x}}$

This is a linear ODE.

$(x^2+1) y' = \frac {x}{y-1}$ $\text {General Solution: } \ \ (y-1)^2 = \answer {\ln (x^2 + 1)} + C$ (For definiteness, the function you enter for your answer should equal $0$ at $x=0$ . If that’s not the case, you might need to rewrite your solution and redefine the constant $C$ .)

This is a separable ODE.

Solve the initial value problem $y' = \cos ^2 x \cos ^2 2y \text { with } y(0) = 0.$ $\text {Solution: } \ \ y = \answer {\frac {1}{2} \arctan \left ( x + \frac {1}{2} \sin 2x \right )}$

This is a separable ODE.

Solve the initial value problem $y' + (\tan x) y = \sec x \text { with } y(0) = -3.$ $\text {Solution: } \ \ y = \answer {\sin x - 3\cos x}$

This is a linear ODE. The integral of $\tan x$ is $-\ln |\cos x|$ .

Sample Quiz Questions

Let $y(x)$ be the solution to the initial value problem $\frac {dy}{dx} = -(1 + 3 x^2)y^2$ and $y(0) = 1/2$ . What is the value of $y(1)$ ?

$\displaystyle \frac {1}{6}$ $\displaystyle \frac {1}{4}$ $\displaystyle \frac {1}{3}$ $\displaystyle \frac {1}{2}$ $\displaystyle \frac {\pi }{4}$ $\displaystyle 1$

This is a separable ODE. Moving all functions of $y$ to the left-hand side and all functions of $x$ to the right-hand side and integrating gives $\int \frac { -1 }{ y^2} ~ dy = \int (1 + 3 x^2) ~ dx,$ which yields $\frac {1}{y} = x^3 + x + C.$ Evaluating at $x = 0$ and $y = 1/2$ gives $2 = 0 + C$ , so $\frac {1}{y} = x^3 + x + 2,$ i.e., $y = \frac {1}{x^3 + x + 2}.$ Plugging in $x = 1$ gives $y = 1/4$ .

Sample Exam Questions

The solution of the initial value problem $\displaystyle x \frac {dy}{dx} + 3y = 7 x^4$ , $y(1) = 1$ , satisfies $y(2) =$

$0$ $1$ $2$ $4$ $8$ $16$

The solution of the initial value problem $\displaystyle \frac {dy}{dx} - 20 x^4 e^{-y} = 0$ , $y(0) = 0$ , satisfies $y(1) =$

$\ln 5$ $\ln 4$ $\ln 3$ $\ln 2$ $1$ $0$

Let $y(x)$ be the solution of the initial value problem $x \frac {dy}{dx} = e^x - y \ \text { with } \ y(\ln 2) = 0.$ Find $y(1)$ .

$\displaystyle \frac {e^2}{2}$ $\displaystyle 2e^2$ $\displaystyle \frac {e}{2}$ $0$ $e-2$ $1$

Let $y(x)$ be the solution of the initial value problem $x \frac {dy}{dx} = y + x^2 \sin x \ \text { with } \ y (\pi ) = 0.$ What is $y(2 \pi )$ ?

$-\pi$ $-2\pi$ $-4\pi$ $0$ $2\pi$ $4\pi$

Consider the initial value problem $(1 + x^2) \frac {dy}{dx} = 2y \ \text { with } \ y(0) = 2.$ What is $\displaystyle \lim _{x \rightarrow \infty } y(x)$ ?

$2e^{\pi }$ $2e^{\pi /2}$ $2e^{\pi /4}$ $1$ $0$ $e^{\pi }$