Originally Posted by
mandy123 Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)
(x^2)y'' + 3xy'=2
This is what I have but came up with the wrong answer
can someone explain what I need to do (Thanks)
y is missing
Let p= y'= dy/dx then let p'=y''=dp/dx
(x^2) dp/dx +3xp= 2
dp/dx + (3/x)p = (2/x^2)
Now i used the linear first-order equation and found
Dx + (3x^2)p = 2x
I then got
(3x^2)p = x^2 + C
Dividing by (3x^2) I got
p= (1/3) + C/(3x^2)
This is not the right answer, the right answer is ln(x) + Ax^-2 + B
I do not have a clue how they got this can someone please help me???
Thanks