Second Order Inhomogeneous ODE.

[Je m'appelle]

Homework Statement

Solve the following equation

f'' - 3f' + 2f = 3

For

f(0) = 0,\ and\ f'(0) = 1

Homework Equations

None.

The Attempt at a Solution

Ok, so, I know how to solve inhomogeneous ODEs when it's on this form

af'' + bf' + cf = g

Where g is a given function.

But how do I proceed when I find a constant instead of a function on the right hand side of the equation?

I'm really confused here.

 

[LCKurtz]

If you really must have an x on the right side, think of your function as 3 + 0x.

In other words, work it the same way.

 

[Je m'appelle]

Ok, so I saw this example in the book

f'' + f' - 2f = x^2

And then the particular solution f_p was found by the following process

As\ g(x) = x^2\ is\ a\ second\ degree\ polynomial\ we'll\ look\ for\ a\ solution\ like

f_p = Ax^2 + Bx + C

Then

\frac{df_p}{dx} = 2Ax + B,\ \frac{d^2 f_p}{dx^2} = 2A

We'll plug-in the derivatives of f_p in the original equation

f'' + f' - 2f = x^2

(2A) + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2

-2Ax^2 + x(2A - 2B) + (2A + B - 2C) = x^2 + 0x + 0

Then

A = \frac{-1}{2},\ B = \frac{-1}{2},\ C = \frac{-3}{4}

So,

f_p = \frac{-x^2}{2} - \frac{x}{2} - \frac{3}{4}

And I tried this same method in my original post, by using 3 = 3x^0, but it did not work.

Please someone help.

 

[Mark44]

You are making this much more complicated than it needs to be. Try y_p = C. Don't forget that you need to solve the homogeneous problem f'' - 3f' + 2f = 0. The solution to the homogeneous problem, plus the particular solution, will be your general solution.

 

[Je m'appelle]

You are making this much more complicated than it needs to be. Try y_p = C. Don't forget that you need to solve the homogeneous problem f'' - 3f' + 2f = 0. The solution to the homogeneous problem, plus the particular solution, will be your general solution.

Ok, so, by considering f_p = C just as you said, I got f_p = \frac{3}{2}, and it worked!

Then the general solution will be:

f(x) = A_1e^{x} + A_2e^{2x} + f_p

f(x) = A_1e^{x} + A_2e^{2x} + \frac{3}{2}

I won't bother you folks showing all the work done to find A_1 and A_2, but I eventually got to

A_1 = -4,\ and\ A_2 = \frac{5}{2}

f(x) = -4e^{x} + \frac{5}{2}e^{2x} + \frac{3}{2}

Let's verify it by plugging-in the above function in the original equation,

f'' - 3f' + 2f = 3

Where,

\frac{df(x)}{dx} = -4e^x + 5e^{2x}

\frac{d^2 f(x)}{dx^2} = -4e^x + 10e^{2x}

(-4e^x + 10e^{2x}) - 3(-4e^x + 5e^{2x}) + 2(-4e^{x} + \frac{5}{2}e^{2x} + \frac{3}{2}) = 3

-4e^x + 10e^{2x} + 12e^x - 15e^{2x} - 8e^x + 5e^{2x} + 3 = 3

(-4e^x + 12e^x - 8e^x) + (10e^{2x} - 15e^{2x} + 5e^{2x}) + 3 = 3

Indeed, 3 = 3

This is pathetically easy, my god.. I don't even know how or why I complicated it so much.

Thank you very much for your time LCKurtz and Mark44.

 

[Mark44]

They can get much more complicated in a hurry. The best practice is to solve the homogeneous problem first, since the homogeneous solutions can impact your choices for your particular solution.

For example, if the equation had been y'' - y' = 3, your particular solution would NOT have been a constant.