Let y = y1(x) be a solution of (H): y'' + p(x)y' + q(x)y = 0 where p and q are continuous function on an interval I. Let a be in I and assume that y1(x) does not =0 on I.

Set y2(x) = y1(x) integral [(e^-integral p(u)du)/(y1^2(t))] dt.

Show that y2 is a solution of (H) and that y1 and y2 are linearly independent.

Show that

Question

Let y = y1(x) be a solution of (H): y'' + p(x)y' + q(x)y = 0 where p and q are continuous function on an interval I.  Let a be in I and assume that y1(x) does not =0 on I.

Set y2(x) = y1(x) integral [(e^-integral p(u)du)/(y1^2(t))] dt.

Show that y2 is a solution of (H) and that y1 and y2 are linearly independent.

Show that

anonymous · Answer

Would you by chance be able to help with this??

@zepdrix

zepdrix · Answer

Hmm I vaguely remember that formula.
Reduction of Order I believe it's called. Used for finding a second solution when you already have a first.

Hmmmmm...

anonymous · Answer

I have not heard of that... I will google, but I am at a total lose on what to do

sirm3d · Answer

suppose there exists some function $v(t)$ such that $y(t)=v(t) y_1(t)$ is a solution to the DE.

This is the reduction of order method.

sirm3d · Answer

put $y=vy_1$
$$\large{y'=v'y_1+vy_1'\y''=v''y_1+2v'y_1'+vy_1''\0=y''+py'+qy=(v''+2v'y_1'+vy_1'')+p(v'y_1+vy_1')+q(vy_1)\=v''+2v'y_1'+pv'y_1+v(y_1''+py_1'+qy_1)\=v''+(2y_1'+py_1)v'+0}$$

anonymous · Answer

so where is y2? or are you saying that y is y2?