# Second order differential equations

• Nov 8th 2008, 03:14 AM
roshanhero
Second order differential equations
What should I try for the particular integral inorder to solve these problems where D means d/dx.
$1.(D^2-4D+4)y=x^3e^{2x}$
$2.\frac{d^2y}{dx^2}+4y=sin^2x$
• Nov 8th 2008, 05:43 AM
Peritus
• Nov 8th 2008, 06:02 AM
roshanhero
But I couldnot get it still.
• Nov 8th 2008, 06:29 AM
Chris L T521
Quote:

Originally Posted by roshanhero
What should I try for the particular integral inorder to solve these problems where D means d/dx.
$1.(D^2-4D+4)y=x^3e^{2x}$

Are you familiar with the so-called "Annihilator" approach?? To see how you do it this way, look at post #7 in my differential equations tutorial [also look at post #6 and #8 for two other techniques].

Going along with this, we solve the homogeneous solution first:

You should end up with $y_c=c_1e^{2x}+c_2xe^{2x}$

Now for the particular solution, solve the non homogeneous equation.

This will lead us to finding the annihilator of $x^3e^{2x}$, which is $\left(D-2\right)^4$

The DE becomes $\left(D-2\right)^2\left(D-2\right)^4=0\implies (r-2)^6=0\implies r=2$ with multiplicity six. Thus, the particular solution with only consist of r repeating 4 times, since 2 were used in the homogeneous solution.

Thus, $y_p=Ax^2e^{2x}+Bx^3e^{2x}+Cx^4e^{2x}+Dx^5e^{2x}$

Now substitute $y_p$ into the original DE and find the coefficients...

Quote:

$2.\frac{d^2y}{dx^2}+4y=sin^2x$
If you still do the annihilator approach, $\frac{d^2y}{dx^2}+4y=\sin^2x\implies \left(D^2+4\right)y=\tfrac{1}{2}-\tfrac{1}{2}\cos (2x)$

Try to do something similar to what I have done above.

I hope this helps.

--Chris