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