Ordinary Differential Equations/Exact 2

First Order Differential Equations

This page details a method for finding the solutions to equations of the form

${\frac {dy}{dx}}+P(x)y=Q(x),$

Using the Integrating Factor: $I(x)=e^{\int P(x)dx}$

Example 1

Consider the following equation:

${\frac {dy}{dx}}+3y=xe^{-3x}+1$

Now the $P(x)=3$ So the integrating factor is:

$I(x)=e^{\int P(x)dx}$

$I(x)=e^{\int 3dx}$

$I(x)=e^{3x}$

Multiply the original equation by $I(x)$

$e^{3x}{\frac {dy}{dx}}+e^{3x}3y=e^{3x}(xe^{-3x}+1)$

$e^{3x}{\frac {dy}{dx}}+e^{3x}3y=x+e^{3x}$

Then try: ${\frac {d}{dx}}(ye^{3x})=e^{3x}{\frac {dy}{dx}}+e^{3x}3y$

Which is equal to the LHS giving us:

${\frac {d}{dx}}(ye^{3x})=x+e^{3x}$

Then integrate with respect to x:

$\int {\frac {d}{dx}}(ye^{3x})dx=\int xdx+\int e^{3x}dx$

Giving:

$ye^{3x}={\frac {x^{2}}{2}}+{\frac {e^{3x}}{3}}+C$

$y={\frac {x^{2}}{2e^{3x}}}+{\frac {1}{3}}+Ce^{-3x}$ (This is the general solution)