Solve Linear Inhomogeneous 2nd Order ODE - Alvin's Question on Yahoo Answers

[MarkFL]

Here is the question:

Solve the given nonhomogeneous ODE by variation of parameters or undetermined coefficients. Give a general sol?

Solve the given nonhomogeneous ODE by variation of parameters or undetermined coefficients. Give a general solution.
Please show work so I can learn. Thanks!

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I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Alvin,

We are given to solve:

\(\displaystyle y''+y=\csc(x)\)

I will use variation of parameters because I don't see how the annihilate the right side of the equation.

We see that a fundamental solution set for the corresponding homogeneous equation are:

\(\displaystyle y_1(x)=\cos(x)\)

\(\displaystyle y_2(x)=\sin(x)\)

And so we take as our particular solution:

\(\displaystyle y_p(x)=v_1(x)\cos(x)+v_2(x)\sin(x)\)

Next, we want to solve the system:

\(\displaystyle \cos(x)v_1'+\sin(x)v_2'=0\)

\(\displaystyle -\sin(x)v_1'+\cos(x)v_2'=\csc(x)\)

From the first equation, we find:

\(\displaystyle v_1'=-\tan(x)v_2'\)

And so substituting into the second equation, we obtain:

\(\displaystyle \frac{\sin^2(x)}{\cos(x)}v_2'+\cos(x)v_2'=\csc(x)\)

Multiply through by $\cos(x)$:

\(\displaystyle \sin^2(x)v_2'+\cos^2(x)v_2'=\cot(x)\)

Apply a Pythagorean identity on the left:

\(\displaystyle v_2'=\cot(x)\)

Hence, this implies:

\(\displaystyle v_1'=-1\)

Integrating, we obtain:

\(\displaystyle v_1=-x\)

\(\displaystyle v_2=\ln|\sin(x)|\)

And so out particular solution is:

\(\displaystyle y_p(x)=-x\cos(x)+\sin(x)\ln|\sin(x)|\)

And then by superposition, we obtain the general solution:

\(\displaystyle y(x)=y_h(x)+y_p(x)=A\cos(x)+B\sin(x)-x\cos(x)+\sin(x)\ln|\sin(x)|\)