Integrating Factor and $\frac{dy}{dx}+y=3x$

Question

$\newcommand{\diff}{\frac{dy}{dx}}$

I was given the following problem: $$\diff+y = e^{3x}$$

My approach

The integrating factor I found to be $P(x)=1$. Then my method to tackle the problem resulted in the following: $$e^{\int P(x)dx}=e^{\int 1dx}=e^x$$

I began to end the problem in these steps: $$ \begin{align} \diff+y &= e^{3x} \\ e^x\diff+e^xy &= e^{4x} \\ \diff\left[e^xy\right] &= e^{4x} \\ \int\diff\left[e^xy\right]dx &= \int e^{4x}dx \\ e^xy &= \frac{1}4e^{4x}+C \\ y &= \frac{1}4e^{3x}+Ce^{-x} \end{align}$$

Is my result correct?

Answer 1

Looks good to me. For the future, note a couple of sanity checks:

• it satisfies the original equation
• your equation is linear and 1st degree, so I expect one family of solutions ($Ce^{-x}$ here) and one particular solution $e^{3x}/4$ here

You did make a typo on the first step writing a $-$ instead of a $+$, but you followed by using $+$ instead of $-$...