# Find the second solution of a differential equation

• Jun 24th 2008, 07:52 AM
kithy
Find the second solution of a differential equation
Need help...Given a differential equation and its first solution y1 find the second solution y2 using reduction of order.

x^2 y'' - 3xy' +5y =0; y1 = x^2 cos (ln x)

thank you so much!
• Jun 24th 2008, 08:10 AM
topsquark
Quote:

Originally Posted by kithy
Need help...Given a differential equation and its first solution y1 find the second solution y2 using reduction of order.

x^2 y'' - 3xy' +5y =0; y1 = x^2 cos (ln x)

thank you so much!

Let $\displaystyle y_2(x) = v(x) \cdot y_1(x)$

Then
$\displaystyle y_2' = v'y_1 + vy_1'$

$\displaystyle y_2'' = v''y_1 + 2v'y_1 + vy_1''$

$\displaystyle x^2(v''y_1 + 2v'y_1 + vy_1'') -3x(v'y_1 + vy_1') + 5(vy_1) = 0$

Now look at the term proportional to v(x):
$\displaystyle y_1'' - 3xy' + 5y_1$

But this is just 0 according to the differential equation. So we get
$\displaystyle x^2y_1v'' + (2x^2y' - 3xy)v' = 0$

and you can take it from here.

-Dan