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

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)