[SOLVED]-9.1.98 solve DE

karush

Well-known member
$\tiny [9.1.98]$ solve
123
$\begin{array}{rl} e^{5x}\,\frac{dy}{dx} + 5e^{5x}y &= e^{4x} \\ \dfrac{d}{dx}\left( e^{5x}y\right) &=e^{4x} \\ e^{5x}y&= \int{ e^{4x}dx} \\ e^{5x}y &= \frac{1}{4}e^{4x} + C \\ y &= \frac{1}{4}\,{e}^{-x} + C\,{e}^{-5\,x} \end{array}$
just reviewing some problems before Sept classes start up
I think this ok not sure how to check it with W|A
possible typos
Mahalo

Last edited:

MHB Math Helper

County Boy

Member
Very good! Here is another way to solve that equation (how I would probably have done it myself):
The equation is $e^{5x}\frac{dy}{dx}+ 5e^{5x}y= e^{4x}$.​
Divide through by $e^{5x}$ to get​
$\frac{dy}{dx}+ 5y= e^{-x}$.​
That is a "linear differential equation with constant coefficients". It's homogeneous part is $\frac{dy}{dx}+ 5y= 0$ which is the same as $\frac{dy}{dx}= -5y$ and then, in "differential" form, $\frac{dy}{y}= -5dx$. Integrating both sides, $ln(y)= -5x+ C$. Taking the expontial of both sides, $y= C'e^{-5x}$ where $C'= e^C$ is just another constant. That is the general solution to the "associated homogeneous equation".​
Now to find a solution to the entire equation, since "$e^{-x}$ is a type of function we get as solution to a "linear differential equation with constant coefficients", we try a solution of the form $y= Ae^{-x}$. Then $y'= -Ae^{-x}$ and the equation becomes $-Ae^{-x}+ 5Ae^{-x}= 4Ae^{-x}= e^{-x}$. Since $e^{-x}$ is never 0 we can divide by it to get 4A=1 so $A= \frac{1}{4}$. $\frac{e^{-x}}{4}$ is a solution to the full equation so the general solution to the full equation is $y(x)= C'e^{-5x}+ \frac{e^{-x}}{4}$.

Mahalo