Solving Second Oder IVP

Question

I am trying to figure out a way to solve the following IVP;

$y’’+y’+y=\sin^2t$

I am just really stuck on what to do with the $\sin^2t$ so any help would be so much appreciated thank you!

Answer 1

First Solve the Homogeneous Equation $${y''+y'+y=0}$$ the solution have the form $y=e^{mt}$; then we have ${m^2+m+1}$ solve for m $${m_{12}={-1\ \pm\sqrt{1-4(1)(1)}\over 2(1)}}$$ $${m_{12}={-{1\over 2}\pm{\sqrt3\over 2}i}}$$ $${y=e^{m_{12}t}\rightarrow{e^{\left({-{1\over 2}\ \pm \ {{\sqrt3\over 2}}i}\right)t}}}$$ $${e^{\left({-{t\over 2}\ \pm\ {{\sqrt3\over2}\it}}\right)}=e^{{-{t\over 2}\ }}e^{\pm \left({\{\sqrt3\over2}t\}\right) i}}$$ Recall Euler's Formula $\star\ \mathbf{e^{\pm\space i\theta}=cos(\theta)\pm\space i\ sin(\theta)}.$ Finally we have our Homogeneous Solution $$\mathbf{y_{h}(t)=e^{{-{t\over 2}\ }}\left[C_{1}cos\left({\sqrt3\over2}t\right)+C_{2}sin\left({\sqrt3\over2}t\right)\right]}$$

For the Particular Solution $y_{p}=sin^2(t)$ replace it with $y_{p}={1+cos(2t)\over2}$. By Undeterminated Coefficientes our has the form $$y_{p}= A+B\ cos(2t)+C\ sin(2t)$$ $$y'_{p}=2C\ cos(2t)-2B\ sin(2t)$$ $$y''_{p}= -4B\ cos(2t)-4C\ sin(2t)$$

Replace all in the original ODE

$$A+(2C-3B)\ cos(2t)-(2B+3C)\ sin(2t)={1\over2}-\frac 12\ cos(2t)$$

Therefore $A=\frac 12;\ \ \ 2C-3B=-\frac 12;\ \ \ 2B+3C=0\ $ $$A=\frac 12$$ $$B=\frac {3}{26}$$ $$C=-\frac {1}{13}$$

$$\mathbf{y_{p}(t)= \frac 12+\frac {3}{26}\ cos(2t)-\frac {1}{13}\ sin(2t)}$$

Our final Answer

$$\mathbf{y(t)=e^{{-{t\over 2}\ }}\left[C_{1}cos\left({\sqrt3\over2}t\right)+C_{2}sin\left({\sqrt3\over2}t\right)\right]+\frac 12+\frac {3}{26}\ cos(2t)-\frac {1}{13}\ sin(2t)}$$ OR $$\mathbf{y(t)=e^{{-{t\over 2}\ }}\left[C_{1}cos\left({\sqrt3\over2}t\right)+C_{2}sin\left({\sqrt3\over2}t\right)\right]+\frac {1}{26}{\left[3cos(2t)-2sin(2t)+13\right]}}$$

Please tell me if something it's wrong.