Let y=y(x) solve dydx+2y=f(x), f(x)=1 for x[0,1], 0 otherwise, y(0)=0.

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}+2y=f(x)$, $f(x)=1$ for $x\in[0,1]$, $0$ otherwise, $y(0)=0$. Then $y\left(\dfrac32\right)$ is:

Accepted Answer

On $[0,1]$: $y=\dfrac12-\dfrac12e^{-2x}$; $y(1)=\dfrac{e^2-1}{2e^2}$.

On $x>1$: $y=Ce^{-2x}$; matching at $x=1$: $C=\dfrac{e^2-1}{2}$.

$$y\left(\frac32\right) = \frac{e^2-1}{2}e^{-3} = \frac{e^2-1}{2e^3}.$$

Solutions · 1