# Solve differential equation $y''+y=\cos^2x$

Solve differential equation $y''+y=\cos^2x$

We are looking at a homogeneous equation:

$$y''+y=0$$ $$\lambda^2+1=0\Rightarrow \lambda_1=-i,\lambda_2=i\Rightarrow y_h=C_1e^{0x}\cos x+C_2e^{0x}\sin x=C_1\cos x+C_2\sin x$$

Now we find a particular solution: $$y''+y=\cos^2x=\frac{1}{2}+\cos(2x)$$

$$y''+y=\frac{1}{2}$$

$$y_{p1}=1/2$$ $$y''+y=\cos(2x)$$ $$y_{p2}=A\sin 2x+B\cos 2x,y_{p2}''=-4A\sin 2x-4B\cos 2x$$ $$\Rightarrow A=0,B=-1/3, y_{p2}=-\frac{1}{3}\cos 2x$$ $$\Rightarrow y=y_h+y_{p1}+y_{p2}=C_1\cos x+C_2\sin x+\frac{1}{2}-\frac{1}{3}\cos 2x$$

Is this correct?

• set this in your equation and differentiate it – Dr. Sonnhard Graubner Feb 11 '17 at 8:45
• If you correct the trigonometric formula to $\cos^2x=\frac12+\frac12\cos(2x)$, this could look correct. – Lutz Lehmann Feb 11 '17 at 8:48
• it should be $$y(x)=c_2 \sin (x)+c_1 \cos (x)-\frac{1}{6} \cos (2 x)+\frac{1}{2}$$ – Dr. Sonnhard Graubner Feb 11 '17 at 9:01

As @LutzL said in the comments, use:

$$\cos^2\left(x\right)=\frac{1+\cos\left(2x\right)}{2}\tag1$$

And you're correct.

Another approach, use Laplace transform:

$$\mathcal{L}_x\left[\text{y}''\left(x\right)+\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=\mathcal{L}_x\left[\cos^2\left(x\right)\right]_{\left(\text{s}\right)}=\frac{1}{2}\cdot\left(\mathcal{L}_x\left[1\right]_{\left(\text{s}\right)}+\mathcal{L}_x\left[\cos\left(2x\right)\right]_{\left(\text{s}\right)}\right)\tag2$$

So, we use that:

• $$\mathcal{L}_x\left[\text{y}''\left(x\right)\right]_{\left(\text{s}\right)}=\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}'\left(0\right)\tag3$$
• $$\mathcal{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=\text{Y}\left(\text{s}\right)\tag4$$
• $$\mathcal{L}_x\left[1\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\tag5$$
• $$\mathcal{L}_x\left[\cos\left(2x\right)\right]_{\left(\text{s}\right)}=\frac{\text{s}}{4+\text{s}^2}\tag6$$

So, we get:

$$\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}'\left(0\right)+\text{Y}\left(\text{s}\right)=\frac{1}{2}\cdot\left(\frac{1}{\text{s}}+\frac{\text{s}}{4+\text{s}^2}\right)\tag7$$

Solving for $\text{Y}\left(\text{s}\right)$:

$$\text{Y}\left(\text{s}\right)=\frac{\frac{1}{2}\cdot\left(\frac{1}{\text{s}}+\frac{\text{s}}{4+\text{s}^2}\right)+\text{s}\cdot\text{y}\left(0\right)+\text{y}'\left(0\right)}{1+\text{s}^2}\tag8$$

Now, using inverse Laplace transform:

$$\text{y}\left(x\right)=\frac{3-\cos\left(2x\right)+\cos\left(x\right)\left(6\text{y}\left(0\right)-2\right)+6\text{y}'\left(0\right)\sin\left(x\right)}{6}\tag9$$