Ordinary Differential Equations/Exact 2

[[../First Order|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:

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

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 $I(x)$

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

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

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

Which is equal to the LHS giving us:

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

Then integrate with respect to x:

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

Giving:

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

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