Differential Equations/Exact 1

From Differential Equations

First Order Differential Equations

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

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

Exact Differential Equations

Consider the following DE:

$y+x\frac{dy}{dx}=x^2$

This is not separable variable, so we cannot use our previous method. However, note that:

$\frac{d}{dx}xy$	$=x\frac{dy}{dx}+y\cdot 1$
	$=y+x\frac{dy}{dx}$

This is the same as the LHS of the first equation. Substituting, we get

$\frac{d}{dx}xy=x^2$

$xy=\frac{x^3}{3}+C$

This is the general solution of the DE above. When a DE has the exact derivative of a product on one side, it is said to be exact.

Converting to an Exact Differential Equation

Consider an equation of the form

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

where P(x), Q(x) and y are all functions of x. This is a first-order linear differential equation. For this to work this form must be closely adhered to - the derivative must be by itself.

In general these equations are not exact. They can, however, be made exact by multiplying through by an integrating factor, I(x), another function of x.

Multiply our original equation by I(x):

(1):$I(x)\frac{dy}{dx}+I(x)P(x)y=I(x)Q(x)$

This will be our new, solvable, DE. Now consider the derivative of the product below:

(2):$\frac{d}{dx}(I(x)\cdot y)=I(x)\frac{dy}{dx}+I^{\prime }(x)y$

Now, if we make the RHS of (1) equal to the LHS of (2), then

(3):$I(x)Q(x)=\frac{d}{dx}(I(x)\cdot y)$

Which, by the other halves of the equations, makes:

(4):$I(x)\frac{dy}{dx}+I(x)P(x)y=I(x)\frac{dy}{dx}+I^{\prime }(x)y$

Which simplifies to:

(5):$I(x)P(x)=I^{\prime }(x)$

By equating the equations to get (3) forces multiplying by I(x) to produce a derivative of a product on the RHS of (1), i.e.

(6):$I(x)\frac{dy}{dx}+I(x)P(x)y=\frac{d}{dx}(I(x)\cdot y)$

The new DE is therefore exact, and can be solved more easily. We now find the function I(x) from (5). We will change notation slightly here.

$\frac{dI}{dx}=I\cdot P(x)$

$\frac{1}{I}\frac{dI}{dx}=P(x)$

$\int \frac{1}{I}dI=\int P(x)dx$

$\ln |I|=\int P(x)dx$

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

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

We take this to be our integrating factor. We can ignore the negative factor, because when both sides of the DE are multiplied by it, they will cancel. So, our integrating factor is:

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

To solve the DE, we then multiply by this factor, and solve the equation, given that one side will be able to be turned into a derivative of a product.