Originally Posted by

**ThePerfectHacker** You have,

$\displaystyle \frac{dy}{dx}-(.2)y=-(.02)x$

This is a *Linear-First Order Differencial Equation*

Thus,

$\displaystyle P(x)=-.2$ and $\displaystyle Q(x)=-.02x$

Thus,

$\displaystyle I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}$

Thus, the solutions are,

$\displaystyle y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)$

Substitute,

$\displaystyle y=e^{.2x}\int (-.02)e^{-.2x}x dx$

Thus,

$\displaystyle y=-(.02)e^{.2x}\int xe^{-.2x}dx$

Integrate by parts,

$\displaystyle y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)$

Thus,

$\displaystyle =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)$

Thus,

$\displaystyle y=.1x+.5+C$

Thus all solutions are,

$\displaystyle y=.1x+C$

Maybe I made a mistake I did not check my work, need to go soon.