<s>[INST] Solving a Cauchy problem, differential equation I have the following Cauchy problem
\begin{cases} y'(x) + \frac{1}{x^2-1}y(x) = \sqrt{x+1} \\ y(0) = 0 \end{cases} 
I proceed by finding $e^{A(x)}  $ where $A(x)$ is the primitive of $a(x)= \frac{1}{x^2-1}$ :
$$\int A(x)dx=\int \frac{1}{x^2-1}dx= \frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)+c  $$
then  I obtain : $$e^{A(x)}=e^{\frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)}=\Big(\frac{|x-1|}{|x+1|}\Big)^{\frac{1}{2}}=\sqrt{\frac{|x-1|}{|x+1|} }$$
I have attempted to solve it in this way:
$$ \sqrt{\frac{|x-1|}{|x+1|} }\cdot y'(x) + \frac{1}{x^2-1}\cdot \sqrt{\frac{|x-1|}{|x+1|} }y = \sqrt{x+1}\cdot\sqrt{\frac{|x-1|}{|x+1|} }$$
$$\sqrt{\frac{|x-1|}{|x+1|} }*y(x) =\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
$$y(x) =\Big(\sqrt{\frac{|x-1|}{|x+1|} }\Big)^{-1}\cdot\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.
 [/INST] Note that since you have on the right side $\sqrt{x+1}$, we may assume that $x\geq -1$. Moreover the coefficient $\frac{1}{x^2-1}$ implies that  $x\not=\pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
$$\sqrt{\frac{1-x}{1+x}}\cdot y(x) =\int \sqrt{1-x}\,dx.$$
Can you take it from here? ... and do not forget the constant of integration!
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