Last modified on 25 November 2010, at 18:34

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 1Edit

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)