Differential Equations
Solution of ODE
MMTS_Full_Test_07
Grade 12
Question:
Let $y=y(x)$ be solution of $\dfrac{dy}{dx}+2y=f(x)$, where $f(x)=\begin{cases}1 & x\in[0,1]\\0 & \text{otherwise}\end{cases}$. If $y(0)=0$, then $y(\ln 2)$ is
$\dfrac{e^2-1}{4e^2}$
$\dfrac{e^2-1}{2e^2}$
$\dfrac{1}{4}$
$\dfrac{3}{8}$
Step-by-Step Solution
Key Concept: Solve the linear ODE in two regions; match at $x=1$
IF$=e^{2x}$. For $x\in[0,1]$: $y=\frac{1}{2}+Ce^{-2x}$. $y(0)=0\Rightarrow C=-1/2$. $y(\ln 2)=\frac{1}{2}-\frac{1}{2}e^{-2\ln 2}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}$.
Correct Answer: 4