Solving second order non homogeneous diff. equation by a particular method

[Telemachus]

Hi. I have to solve this:

y''-3y'+2y=e^x,
Using the replacement y=\phi y_1 being y_1 a solution of the homogeneous differential equation. I can't do it "traditionally", I have to use this method.

So I have to solve that \phi ''y_1+ \phi ' \left[2y_1'+Py_1] \right]=e^x

So, this is what I did:

y_1=C_1 e^x+C_2e^{2x}\rightarrow y_1'=C_1 e^x+2C_2e^{2x}
Then:
\phi ''C_1 e^x+\phi '' C_2 e^{2x}+ \phi ' \left[C_2e^{2x}-C_1e^x \right]=e^x
What should I do from here? I don't know how to handle \phi

Bye there.

 

[rock.freak667]

This looks like a reduction of order type thing, if y₁ is a solution of y''-3y'+2y = 0, why don't you just choose a 'y₁' from the equation and then substitute?

 

[Telemachus]

You're right, that's what it is. But I don't know what you mean with the substitution you proposed. Isn't it what I did? I mean this:
\phi ''C_1 e^x+\phi '' C_2 e^{2x}+ \phi ' \left[C_2e^{2x}-C_1e^x \right]=e^x
What I have to find is \phi ', but I don't know how. May be I shouldn't use the constants, but I'm not sure about it.

Thanks for posting BTW.

This is how I say thanks to you

:D

 

[Mark44]

I think this technique is also called "variation of parameters." If I remember correctly, what you have is y₁ is one of the solutions of the homogeneous equation, y'' - 3y' + 2y = 0.

Let y = φy₁ be a solution of the nonhomogeneous equation y'' - 3y' + 2y = e^x.

y' = φ'y₁ + φy₁'
y'' = φ''y₁ + 2φ'y₁' + φy₁''

When you substitute the right sides of the three equations above into the nonhomogeneous equation, group the terms so that you have φ'' times some expression, φ' times some expression, and φ times some expression.

You should get something like this: φ''(y₁'' - 3y₁' + 2y₁) + <other stuff> = e^x. The quantity in parentheses simplifies to 0, since y₁ is one of the solutions to the homogeneous problem. This reduces the order of the differential equation, which should allow you to find φ, and from it, your other solution to the nonhomogeneous problem.

 

[rock.freak667]

I think this technique is also called "variation of parameters.

Variation of parameters has the Wronskian in it if I remember from like 2 years ago.

This is how I say thanks to you

:D

I like me some Hendrix.

But Mark44 has posted some sound guidance there.

 

[Telemachus]

I got it. I've made a few mistakes here. At first, I shouldn't take the entire solution for the DE when applying the formula, I just needed one solution, y1. The other thing I shouldn't use are the constants. Doing that one gets a DE equation on \phi of second order. Getting the solution and replacing back in y=\phi y_1 gives the entire solution for the original DE.

Thank you both.
Bye there.