Solve y″+s2y=bcos⁡sx where s and b are constants. I have tried undetermined coefficients, but it makes such a mess that I keep getting lost, I also tried variation of parameters. Really my issue is staying organized enough to get the solution, my solutions are always just a little off from the book.I have solved the complementary homogenous equation, this gave me a solution with cos⁡sx in it... So that's off limits in the particular solution. I looked for a solution in xsinsx and xcos⁡sx, but as I said, it was a mess.

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Solve y″+s2y=bcos⁡sx where s and b are constants. I have tried undetermined coefficients, but it makes such a mess that I keep getting lost, I also tried variation of parameters. Really my issue is staying organized enough to get the solution, my solutions are always just a little off from the book.I have solved the complementary homogenous equation, this gave me a solution with cos⁡sx in it... So that&#039;s off limits in the particular solution. I looked for a solution in xsinsx and xcos⁡sx, but as I said, it was a mess.

sukljama2 · Accepted Answer

You need to differentiate eg y=Axcos⁡sx twice using the chain rule.So y′=Acos⁡sx−Asxsin⁡sxAnd y″=−2Assin⁡sx−As2xcos⁡sxThen y″+s2y=−2Assin⁡sxBut you need cos⁡sx and I&#039;m sure you can see how to do that now.Because the equation is linear, if the expressions get messy you can deal with the sin and cos terms from y=Axsin⁡sx+Bxcos⁡sx separately and get the right linear combination at the end. Sometimes it can be easier to convert everything to exponential functions.

Matthew Rodriguez · Accepted Answer

You might find the Second Order Linear Equations sections of my differential equations notes helpful. In this case you start with the complex version of your equation,  P(D)y=beisx where  P(D)=D2+s2=(D+is)(D−is), D being the derivative operator. Here is is one of the two roots of P: we write y=eisxu, and by the Exponential Shift Theorem P(D)eisxu=eisxP(D+is)u=eisx(D+2is)Du. With v=Du we now want (D+2is)v=b. That has a constant solution v=b2is=−ib2s. Integrating, u=−ibx2s (we can disregard the arbitrary constant, which leads to a solution of the homogeneous equation). So y=−ibx2seisx is a solution of y+s2y=beisx. Writing eisx=cos⁡(sx)+isin⁡(sx) and taking the real part, y=bx2ssin⁡(sx) is a solution of your differential equation.

alenahelenash · Accepted Answer

Here is an ad-hoc method to solve this (which was derived from my answer here: What is the quickest way to solve this 2nd Order Linear ODE?) First notice that(y′cos⁡sx+sysin⁡sx)′=y″cos⁡sx−sy′sin⁡x+sy′sin⁡sx+s2ycos⁡sx=cos⁡sx(y″+s2y)Thus (y′cos⁡sx+sysin⁡sx)=∫bcos2⁡sx+ANow (y′cos⁡sx+sysin⁡sx)=(ycos⁡sx)′cos2⁡sxThus ycos⁡sx=∫sec2⁡sx(∫bcos2⁡sx+A)+BFor a more general method, you can also try Laplace Transform (it might be easier to bring in complex numbers first, as in D Lim&#039;s answer). The method in D Lim&#039;s answer is what I would recommend reading first, though.

Solve y''+s^{2}y=b \cos sx where s and b are constants. I have tried undetermined coefficients

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