Question

Differential Equations Problem.
Can someone help me just through the steps of solving this DE
x^2y''-3xy'+4y=x^2ln(x) I need the general solution

Answer 1

First I found the Yc (complementary solution using Caushy Euler Equation and i got yc=C1.e^2x + C2.xe^2x and i just dont know what to do next (usually i solve it using the variation of parameter) but this time it doesnt seem to work! i think i need to substitute x=e^t but i am unable to figure out where to and when.

Answer 2

ok I will do this on paper and then attach it, one second.

Answer 3

Thank you :)

Answer 4

\[
A x^2+2 B x^2 \ln
(x)+\frac{1}{6} x^2 \ln
^3(x)
\]

Answer 5

and what is that?

Answer 6

That is the general solution.
You first try to find a general solution for
x^2y''-3xy'+4y=0
in the form y=x^r
you find that r= 2 is a double root.
and you continue from there.

Answer 7

yes that what i did (i said that i found the complementary solution using Cauchy Euler Equation ) which is exactly like this! but what did you do after that?

Answer 8

the cauchy euler solution that you got is wrong.. just take a look at the solution i did, you also have to use variation of parameters. ( in my solution the homogenous solution is the comlemetary solution). Good luck.

Answer 9

what I just did is the correct method, I am not sure if I did any calculation mistakes.. so you should go over my solution and redo it.

Answer 10

Thank you very much! but it is the same cauchy euler solution double root =2 again thank you

Answer 11

ohh you are right it is wrong! i left out the ln

Answer 12

ok no problem!

Answer 13

Sir you are a boss!

Answer 14

After finding that x^2 is a solution of the homegenous equation, you do the method of
reduction of order by trying to find a solution of the
form y= v(x) x^2 and you obtain after plugging it into the equation
x^4 v''(x)+x^3 v'(x)=0

x v''(x)+ v'(x)=0
you find v'(x) =K/x
so v(x) =ln(x)
Hence the second solution
is x^2 ln (x)
Now the general solution for the homogenous equation is
A x^2 + B x^2 ln(x)