Question

solve the differential equation.

6y = e^{x/6}+y

the solution is y =

Explanation:

Step1: Rewrite the differential equation

First, rewrite the given differential equation \(6y'=e^{x/6}+y\) in the standard linear form \(y' + P(x)y = Q(x)\).
Divide the entire equation by \(6\): \(y'-\frac{1}{6}y=\frac{1}{6}e^{x/6}\). Here \(P(x)=-\frac{1}{6}\) and \(Q(x)=\frac{1}{6}e^{x/6}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
Calculate \(\int P(x)dx=\int-\frac{1}{6}dx=-\frac{1}{6}x\). So \(\mu(x)=e^{-\frac{1}{6}x}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(y'-\frac{1}{6}y=\frac{1}{6}e^{x/6}\) by \(e^{-\frac{1}{6}x}\):
\(e^{-\frac{1}{6}x}y'-\frac{1}{6}e^{-\frac{1}{6}x}y=\frac{1}{6}\).
The left - hand side is the derivative of the product \((e^{-\frac{1}{6}x}y)\) by the product rule \((uv)' = u'v+uv'\) (where \(u = e^{-\frac{1}{6}x}\) and \(v = y\), \(u'=-\frac{1}{6}e^{-\frac{1}{6}x}\)).

Step4: Integrate both sides

Integrate \((e^{-\frac{1}{6}x}y)'=\frac{1}{6}\) with respect to \(x\).
\(\int(e^{-\frac{1}{6}x}y)'dx=\int\frac{1}{6}dx\).
We get \(e^{-\frac{1}{6}x}y=\frac{1}{6}x + C\).

Step5: Solve for \(y\)

Multiply both sides by \(e^{\frac{1}{6}x}\) to isolate \(y\):
\(y = e^{\frac{1}{6}x}(\frac{1}{6}x + C)\).

Answer:

\(y = e^{\frac{x}{6}}(\frac{x}{6}+C)\)