Solving Ordinary Differential Equation

[jenn9580]

I need to solve the following DE that satisfies the initial condition of y(4)=5.

(x^2-9)y' + xy = 2x(x^2-9)

I started by finding that the integrating factor is x/(x^2-9).

I have the solution from Maple but I am not sure how to get there.

 

[galactus]

$\displaystyle (x^{2}-9)y'+xy=2x(x^{2}-9)$

Divide by $\displaystyle (x^{2}-9)$:

$\displaystyle y'=\frac{x}{x^{2}-9}y=2x$

The integating factor is $\displaystyle e^{\int \frac{x}{x^{2}-9}dx}=\sqrt{x^{2}-9}$

Multiplying through by the I.F., we get:

$\displaystyle \frac{d}{dx}[y\sqrt{x^{2}-9}]=2x\sqrt{x^{2}-9}$

Integrate. The left side cancels the derivative because we have the anti-derivative of a derivative:

$\displaystyle y\sqrt{x^{2}-9}=\frac{2}{3}(x^{2}-9)^{\frac{3}{2}}+C$

Now, solve for y and use the initial condition to find C. Then, you're done.

 

[jenn9580]

Thank you! My problem was in integrating & seeing that the terms on the left turn into the I.C. times y. I think I get it now. Thanks again!