Thread: 2nd Order ODE

1. 2nd Order ODE

Can anyone take me through solving this 2nd Order ODE?

$\displaystyle y'' - 4y' + 3y = e^3^x$

Thanks!

2. Re: 2nd Order ODE Originally Posted by remember $\displaystyle y'' - 4y' + 3y = e^3^x$
Just routine knowing the corresponding theory. Have you studied it?

3. Re: 2nd Order ODE

First calculate the homogenous DE:
$\displaystyle y''-4y'+3y=0$ and determine the general solution, afterwards you've to find a particular solution.

Can you do that?

4. Re: 2nd Order ODE

I understand that i must first find a complementary function, which i worked out as:

$\displaystyle y(x)=Ae^x + Be^3^x$ - where A and B are constants.

My problem is with the next step. I've previously done examples where the homogeneous expression is equal to a quadratic expression... what do i do in the case of an exponential?

5. Re: 2nd Order ODE

That looks good for the general solution, now the particular solution will look like: $\displaystyle a\cdot x \cdot e^{3x}$ (because of $\displaystyle e^{3x}$), you have to find $\displaystyle a$. Calculate the first derivative and second derivative of$\displaystyle a\cdot x \cdot e^{3x}$ and substitute in the DE to find $\displaystyle a$.

6. Re: 2nd Order ODE

I have worked out that:

$\displaystyle y = a e^3^x x$
$\displaystyle y' = e^3^x (3x + 1)$
$\displaystyle y'' = 3e^3^x (3x + 2)$

When you say substitute into the DE, do you mean like this:

$\displaystyle 3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x$?

Then expand the brackets and work out a?

Is my final solution in the form:

$\displaystyle y(x) = a x e^3^x + Ae^x + Be^3^x$?

7. Re: 2nd Order ODE Originally Posted by remember I have worked out that:

$\displaystyle y = a e^3^x x$
$\displaystyle y' = e^3^x (3x + 1)$
$\displaystyle y'' = 3e^3^x (3x + 2)$

When you say substitute into the DE, do you mean like this:

$\displaystyle 3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x$?

Then expand the brackets and work out a?

Is my final solution in the form:

$\displaystyle y(x) = a x e^3^x + Ae^x + Be^3^x$?
Yes. Review the theory (method of undetermined coefficients) in your class notes or textbook.

8. Re: 2nd Order ODE

If you calculate $\displaystyle a$:
$\displaystyle y=a\cdot x\cdot e^{3x}$
$\displaystyle y'=a\left(e^{3x}+3x\cdot e^{3x}\right)=a\cdot e^{3x}+3ax \cdot e^{3x}$
$\displaystyle y''=3a\cdot e^{3x}+3a \left(e^{3x}+3x\cdot e^{3x}\right)$
$\displaystyle =6a \cdot e^{3x}+9ax\cdot e^{3x}$

If you substitute this in the equation you get:
$\displaystyle 6a\cdot e^{3x}+9ax\cdot e^{3x}-4\left(a\cdot e^{3x}+3ax\cdot e^{3x}\right)+3\left(ax\cdot e^{3x}\right)=e^{3x}$
$\displaystyle \Leftrightarrow 2a\cdot e^{3x}=e^{3x}$
$\displaystyle \Leftrightarrow a=\frac{1}{2}$

So the solution of the ODE:
$\displaystyle y=A\cdot e^{x}+B\cdot e^{3x}+\frac{1}{2}\cdot x\cdot e^{3x}$

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