Question

Does anyone know how to find the general solution of the following: (x)dy/dx-4y=exp(x)

Answer 1

\[x\frac{dy}{dx}-4y=e^x\]
Divide both sides by x
\[\frac{dy}{dx}-\frac{4y}{x}=\frac{e^x}{x}\]
We have integrating factor of -4/x
\[\Large \mu(x)=e^{\int\limits -\frac{4}{x}}=e^{-4\ln(x)}=\frac{1}{x^4}\]
Multiply all terms by 1/x^4
\[\frac{1}{x^4}\frac{dy}{dx}-\frac{4y}{x}\frac{1}{x^4}=\frac{e^x}{x^5}\]
notice from LHS is a end-product rule, so
\[\frac{d}{dx}(\frac{1}{x^4}y)=\frac{e^x}{x^5}\]
Integrate both sides
\[\frac{y}{x^4}=\int\limits \frac{e^x}{x^5} dx\]
Solve

Answer 2

Thank you so much :D

Answer 3

you're welcome :)