I have the differential equation $$2\frac{dy}{dx}+3y=e^{-2x}-5$$
I have determined that this needs solving using the integrating factor method.
My workings out are in the image provided.
Are my workings out correct so far? My main question is, how will I integrate with respect for $x$?
$$\int e^{3x/2} \times e^{-2x}-\frac{5}{2}e^{3x/2}~~dx+c$$
this again is shown at the bottom of my workings out.
after integrating this I will then be able to solve for $y$.
Thank you
On
$$2y'+3y=e^{-2x}-5$$ $$y'+\dfrac 23y=\dfrac 12 (e^{-2x}-5)$$ $$(ye^{3x/2})'=\dfrac {e^{3x/2}}2 (e^{-2x}-5)$$ $$ye^{3x/2}=\int \dfrac {e^{3x/2}}2 (e^{-2x}-5)dx$$ Don't forget the factor $1/2$: $$ye^{3x/2}= \color {red}{\dfrac 12} \int (e^{-x/2}-5e^{3x/2})dx$$
Now to integrate use this result: $$\int e^{ax}dx=\dfrac 1 a\int \,\,d(e^{ax})= \dfrac {e^{ax}}{a}+K$$
Edit1: Integration gives us: $$ye^{3x/2}= {\dfrac 12} \int (e^{-x/2}-5e^{3x/2})dx$$ $$ye^{3x/2}= {\dfrac 12}(-2e^{-x/2}-\dfrac {10}3e^{3x/2})+K$$ $$ye^{3x/2}= -e^{-x/2}-\dfrac {5}3e^{3x/2}+K$$
$$y=e^{-3x/2}( -e^{-x/2}-\dfrac {5}3e^{3x/2})+Ke^{-3x/2}$$ And finally :
$$\boxed {y= -e^{-2x}-\dfrac {5}3+Ke^{-3x/2}}$$
HINT: $$\int e^{kx}dx=\frac{1}{k}e^{kx}+c$$ where $k$ is any constant. Ask for more detail if you're still stuck :)