second order differential

[zina_6]

I need help solving a second order differential equation.


L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)


Hence solve the equation y''-3y'+2y=3e^x



This is how I started to work out this problem. Can some one help ?


y1=e^x This is what i am assuming to be y1 and y2 ?
y2= Xe^x

then my W = e^2x

Yp = 9x^2e^x/2


Then the general solution Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2

Did I go about this the wrong way ?

Should I have plugged the value of L (e^x) and L (Xe^x)

[HallsofIvy]

You entered this exact problem on another board several days ago.

Here is my answer on that other board:

Because the problem SAYS "L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)",
yes, you should first "have plugged the value of L (e^x) and L (Xe^x)"!

One of the things you will discover is that from that is that xe^x does NOT satisfy the associated homogeneous equation so that Y= c1 e^x + c2 Xe^x is NOT the general solution to the associated homogeneous equation and that
Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2 is NOT the general solution to the entire equation!