Question

Let \(y=y(x)\) be the solution of the differential equation \(\frac{dy}{dx}+2y=f(x)\), where \(f(x)=\begin{cases}1, & x\in[0,1]\\ 0, & \text{otherwise}\end{cases}\) If \(y(0)=0\), then \(y\left(\frac32\right)\) is :

• A. \(\frac{e^2+1}{2e^4}\)
• B. \(\frac{1}{2e}\)
• C. \(\frac{e^2-1}{e^3}\)
• D. \(\frac{e^2-1}{2e^3}\)

Hint:

Solve on \([0,1]\) with \(f=1\) and \(y(0)=0\), then on \((1,\infty)\) with \(f=0\), using continuity of \(y\) at \(x=1\).

Answer 1

The Correct Option is D

On \([0,1]\): \(y'+2y=1\), integrating factor \(e^{2x}\): \(ye^{2x}=\frac{e^{2x}}2+c\). With \(y(0)=0\): \(c=-\frac12\), so \(y=\frac{1-e^{-2x}}{2}\) and \(y(1)=\frac{1-e^{-2}}{2}\).
For \(x>1\): \(y'+2y=0\Rightarrow y=Ae^{-2x}\). Continuity at \(x=1\): \(Ae^{-2}=\frac{1-e^{-2}}{2}\Rightarrow A=\frac{e^2-1}{2}\).
\(y\left(\frac32\right)=\frac{e^2-1}{2}e^{-3}=\frac{e^2-1}{2e^3}\).
Answer: (4) \(\frac{e^2-1}{2e^3}\)