Question

solve the differential equation.

5y = 7e^{x/5} + y

the solution is y =

Explanation:

Step1: Rewrite the differential equation in standard form

The given differential equation \(5y'=7e^{x/5}+y\) can be rewritten as \(y'-\frac{1}{5}y=\frac{7}{5}e^{x/5}\).
Here, \(P(x)=-\frac{1}{5}\) and \(Q(x)=\frac{7}{5}e^{x/5}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
Since \(\int-\frac{1}{5}dx=-\frac{1}{5}x\), then \(\mu(x)=e^{-x/5}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(y'-\frac{1}{5}y=\frac{7}{5}e^{x/5}\) by \(e^{-x/5}\), we get \(e^{-x/5}y'-\frac{1}{5}e^{-x/5}y=\frac{7}{5}\).
The left - hand side is the derivative of the product \((e^{-x/5}y)'\) (by the product rule \((uv)' = u'v+uv'\), where \(u = e^{-x/5}\) and \(v = y\), \(u'=-\frac{1}{5}e^{-x/5}\)).

Step4: Integrate both sides

Integrate \((e^{-x/5}y)'=\frac{7}{5}\) with respect to \(x\).
\(\int(e^{-x/5}y)'dx=\int\frac{7}{5}dx\).
We know that \(\int(e^{-x/5}y)'dx=e^{-x/5}y\) and \(\int\frac{7}{5}dx=\frac{7}{5}x + C\).
So, \(e^{-x/5}y=\frac{7}{5}x + C\).

Step5: Solve for \(y\)

Multiply both sides by \(e^{x/5}\) to get \(y = e^{x/5}(\frac{7}{5}x + C)\).

Answer:

\(y = e^{x/5}(\frac{7}{5}x + C)\)