Question

solve the differential equation.

2 y^{prime}=5 e^{x / 2}+y

the solution is y=

Explanation:

Step1: Rewrite the differential equation

First, rewrite \(2y' = 5e^{x/2}+y\) in the standard linear form \(y'- \frac{1}{2}y=\frac{5}{2}e^{x/2}\). Here \(P(x)=-\frac{1}{2}\) and \(Q(x)=\frac{5}{2}e^{x/2}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\). Calculate \(\int-\frac{1}{2}dx=-\frac{1}{2}x\), so \(\mu(x)=e^{-x/2}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(y'- \frac{1}{2}y=\frac{5}{2}e^{x/2}\) by \(e^{-x/2}\), we get \((ye^{-x/2})'=\frac{5}{2}\).

Step4: Integrate both sides

Integrate \((ye^{-x/2})'=\frac{5}{2}\) with respect to \(x\). \(\int(ye^{-x/2})'dx=\int\frac{5}{2}dx\). Then \(ye^{-x/2}=\frac{5}{2}x + C\).

Step5: Solve for \(y\)

Multiply both sides by \(e^{x/2}\) to solve for \(y\), \(y = e^{x/2}(\frac{5}{2}x + C)=5xe^{x/2}/2+Ce^{x/2}\).

Answer:

\(y = 5xe^{x/2}/2+Ce^{x/2}\)