Question

solve x"+2x'+x=2(t^-2)(e^-t) using variation of parameters. I have solved the homogeneous to be C1e^(-t)+C2te^-t and solved the wronskian how do I find the particular soln for my gen eqn?

Answer 1

I was thinking my particular is y=z1e^(-t)+z2te^-t but i dont know how to solve for z

Answer 2

Particular solution is y = u1 * y1 + u2* y2

y1= e^(-t)
y2= t* e^(-t)

There are two equations (this is a rule)

(u1)' * y1 + (u2)' * y2 = 0

(u1)' * (y1)' + (u2)' * (y2)' = g(x)

(u1)' and (u2)' are derivates of the functions u1 and u2

g(x) is the right hand side of main eq which is given as 2(t^-2)(e^-t)

you should calculate (u1)' and (u2)' from these two equations. (you can use Cramer's rule,which is where you re gonna use Wronskian of y1 and y2)

Then integrate (u1)' and (u2)' and you will get u1 and u2

and write it on equation y = u1 * y1 + u2* y2