Solution to 2nd order ODE using the D operator method with 2 trig terms on RHS

[mitch_1211]

Hey,

I have the DE

y'' -2y' + 3y = xsin(x) + 2cosh(2x)

Using the D operator as D = $\frac{dy}{dx}$ this becomes

(D² -2D +3)y = xsin(x) + 2cosh(2x)

so y_p = $\frac{1}{p(D^2)}$ operating on xsin(x) + 2cosh(2x)

(i think)

So i know if this was say $\frac{1}{p(D^2)}$ operating on sin(x)
i.e (D² -2D +3)y = sin(x)
then y_p = 1/(D² -2D +3) * sin(x)

and you would substitute D² = -($\alpha$)² and proceed to solve.(where alpha is the coefficient of x in the argument of sin, here it is 1)

My question is, I have a term on the RHS which is polynomial times trig; xsin(x) and a trig term which can be treated as an exponential; 2cosh(2x)

I know I can split this into
y_p = 1/(D² -2D +3) * xsin(x) + y_p = 1/(D² -2D +3) * 2cosh(2x)

But I'm unsure of the 'rules' to use here, i.e for a single trig term you swapped D²
for -($\alpha$)²
But what do you do for a poly times a trig and for the cosh function?

Any points would be much appreciated, if all else fails I will have to resort to the method of undetermined coefficients

 

[jvicens]

.

Hi there,

Thank you for your question. It looks like you are on the right track with using the D operator to solve this differential equation. However, I would like to clarify a few points for you.

Firstly, when you have a term like xsin(x) on the right hand side, you can use the rule that D(p(x)y) = p'(x)y + p(x)Dy. So in this case, you can rewrite xsin(x) as sin(x)y' + cos(x)y. Then you can use the D operator to solve each term separately.

Secondly, for the term 2cosh(2x), you can use the fact that cosh(x) = (e^x + e^-x)/2. So 2cosh(2x) can be rewritten as e^2x + e^-2x. Then you can use the D operator to solve each term separately as well.

Overall, the key is to use the D operator to rewrite each term on the right hand side as a sum of terms that are easier to solve. Then you can apply the method of undetermined coefficients to solve for the particular solution.

I hope this helps. Let me know if you have any further questions. Good luck with your problem!