Kristen Stokes

2022-07-07

I am aware that the solution to a homogeneous first order differential equation of the form $\frac{dy}{dx}=p\left(x\right)y$ can be obtained by simply by rearranging to:
$\frac{dy}{y}=p\left(x\right)dx$
Then it is simply a question of integrating both sides and the answer is straightforward. Now what would happen if RHS had a constant, how a can find a particular solution to this case: $\frac{dy}{dx}=p\left(x\right)y+C$
I know that the general solution would be the sum of the homogeneous equation and the particular solution

Kiana Cantu

Simply rearrange :
$\frac{dy}{dx}-yp\left(x\right)=C$
Now multiply both the sides by ${e}^{\int p\left(x\right)dx}$
$\frac{dy}{dx}{e}^{\int p\left(x\right)dx}-yp\left(x\right){e}^{\int p\left(x\right)dx}=C{e}^{\int p\left(x\right)dx}$
This LHS is nothing else :
$\frac{dy}{dx}{e}^{\int p\left(x\right)dx}-yp\left(x\right){e}^{\int p\left(x\right)dx}=\frac{d}{dx}\left(y\cdot {e}^{\int p\left(x\right)dx}\right)$
Now our differential equation becomes :
$\frac{d}{dx}\left(y\cdot {e}^{\int p\left(x\right)dx}\right)=C{e}^{\int p\left(x\right)dx}$
Hence:
$y=\frac{\int C{e}^{\int p\left(x\right)dx}dx}{{e}^{\int p\left(x\right)dx}}$

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