Solving the ODE $\frac{dy}{dx} -5y = e^x$ with $y(0)=-1$ using integrating factors. Solve 
$$\frac{dy}{dx} -5y = e^x$$
Using initial condition $y(0)=-1$.
I calculate that $c=-0.75$, which gives me$$y=-0.25e^x - \frac{0.75}{e^{-5x}}$$
I'm just asking for verification of answer.
thanks to the people for providing the steps and explaining the process despite me only asking for my answer to be verified
 A: Here are the steps
$$ \frac{d}{dx}[y]-5y=e^x $$
Let $\mu(x)=e^{-5\int dx}=e^{-5x}$, so now we have
$$ \frac{d}{dx}[y] e^{-5x}-5ye^{-5x}=e^xe^{-5x} $$
$$ \frac{d}{dx}[y] e^{-5x}+\frac{d}{dx}\left[e^{-5x}\right]y=e^{-4x} $$
$$ \frac{d}{dx}\left[ye^{-5x}\right]=e^{-4x} $$
$$ d\left[ye^{-5x}\right]=e^{-4x}dx $$
$$ \int d\left[ye^{-5x}\right]=\int e^{-4x}dx $$
$$ ye^{-5x}=-\frac14 e^{-4x}+C $$
$$ y=-\frac14 e^{x}+Ce^{5x} $$
Since $y(0)=-1$, then
$$ -1=-\frac14 e^{0}+Ce^{5(0)} $$
$$ -1=-\frac14 +C $$
$$ C=\frac14 -1=-\frac34 $$
Therefore, 
$$ y=-\frac14 e^{x}-\frac34 e^{5x}=-\frac14 e^x\left(3e^{4x}+1\right) $$
A: solve at first the ode $$\frac{dy}{dx}-5y(x)=0$$ and then for the particular solution make the ansatz $$y_p=Ae^{x}$$
A: This is a Linear Differential Equation, if you know how to solve them, it's not difficult to understand that:
$$\newcommand{\p}[0]{\frac{{\rm d}y}{{\rm d}x}}
\newcommand{\e}[1]{{{\rm e}^{#1}}}
\newcommand{\dx}[0]{{\rm d} x}
\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\ct}[0]{\color{grey}{\text{constant}}}
\p-5y=\e x\\
y\e{\int-5\dx}=\int\e{x}\e{\int-5\dx}\dx\\
y\e{-5x}=\int\e{x-5x}\dx=\int e^{-4x}\dx=\frac{1}{-4}(\e{-4x})+\ct\\
y=\e{5x}\b{\ct-\frac{\e{-4x}}{4}}$$
I hope you can manage the constant with boundary values?
