Can anyone take me through solving this 2nd Order ODE?
$\displaystyle y'' - 4y' + 3y = e^3^x$
Thanks!
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?
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 $\displaystyle a$.
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 $?
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:
$\displaystyle y=A\cdot e^{x}+B\cdot e^{3x}+\frac{1}{2}\cdot x\cdot e^{3x}$