Question

solve the differential equation.
7y = 2e^{x/7}+y
the solution is y =

Explanation:

Step1: Rewrite the differential equation

First, rewrite \(7y'=2e^{x/17}+y\) as \(y'-\frac{1}{7}y = \frac{2}{7}e^{x/17}\). This is a first - order linear differential equation of the form \(y'+p(x)y = q(x)\), where \(p(x)=-\frac{1}{7}\) and \(q(x)=\frac{2}{7}e^{x/17}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int p(x)dx}\). Since \(\int-\frac{1}{7}dx=-\frac{1}{7}x\), then \(\mu(x)=e^{-x/7}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(y'-\frac{1}{7}y=\frac{2}{7}e^{x/17}\) by \(e^{-x/7}\), we get \((ye^{-x/7})'=\frac{2}{7}e^{x/17}e^{-x/7}=\frac{2}{7}e^{x/17 - x/7}=\frac{2}{7}e^{- 10x/119}\).

Step4: Integrate both sides

Integrate \((ye^{-x/7})'=\frac{2}{7}e^{-10x/119}\) with respect to \(x\).
\(ye^{-x/7}=\frac{2}{7}\int e^{-10x/119}dx\). Let \(u =-\frac{10x}{119}\), then \(dx=-\frac{119}{10}du\).
\(\int e^{-10x/119}dx=-\frac{119}{10}e^{-10x/119}+C\).
So \(ye^{-x/7}=\frac{2}{7}\times(-\frac{119}{10})e^{-10x/119}+C=-\frac{17}{5}e^{-10x/119}+C\).

Step5: Solve for \(y\)

Multiply both sides by \(e^{x/7}\) to get \(y = Ce^{x/7}-\frac{17}{5}e^{x/7 - 10x/119}=Ce^{x/7}-\frac{17}{5}e^{x/119}\).

Answer:

\(y = Ce^{x/7}-\frac{17}{5}e^{x/119}\)