Ordinary Differential Equations/Exact 2

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:

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

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

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

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:

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

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)