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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. (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. (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. (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.
This is a linear ODE.
(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
This is a separable ODE.
Solve the initial value problem
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 and $y(0) = 1/2$. What is the value of $y(1)$?

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

### 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 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 What is $y(2 \pi )$?

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

Consider the initial value problem What is $\displaystyle \lim _{x \rightarrow \infty } y(x)$?

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