Solve $y-(1+\cos x)y''-(x+\sin x)y'=0,\space y(0)=1,\space y'(0)=0$ $$y-(1+\cos x)y''-(x+\sin x)y'=0,\space y(0)=1,\space y'(0)=0$$
Someone asked me the solution to this DE but I have little background on it, so can't solve it. If anyone can give me some insight, it would be appreciated.
 A: This was to long to put in a comment. You can write your equation in the form of 'Sturm-Lioiville equation':
$$\text{y}\left(x\right)-\left(1+\cos\left(x\right)\right)\cdot\text{y}''\left(x\right)-\left(x+\sin\left(x\right)\right)\cdot\text{y}'\left(x\right)=0\space\Longleftrightarrow\space$$
$$\frac{\text{d}}{\text{d}x}\left\{e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}'\left(x\right)\right\}-\frac{e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}\left(x\right)}{1+\cos\left(x\right)}=0$$
So, we can write:
$$\int\frac{\text{d}}{\text{d}x}\left\{e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}'\left(x\right)\right\}\space\text{d}x=e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}'\left(x\right)=\int\frac{e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}\left(x\right)}{1+\cos\left(x\right)}\space\text{d}x$$
And:
$$\text{y}\left(x\right)=\int\left\{e^{-x\tan\left(\frac{x}{2}\right)}\int\frac{e^{x\tan\left(\frac{x}{2}\right)}\cdot\text{y}\left(x\right)}{1+\cos\left(x\right)}\space\text{d}x\right\}\space\text{d}x$$
