Solving a Cauchy problem, differential equation

Question

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.

Answer 1

Computing $$\mu(x)=e^{\int\frac{1}{x^2-1}dx}=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ then you will get $$\int\frac{d}{dx}\left(\frac{\sqrt{1-x}y(x)}{\sqrt{x+1}}\right)=\int\sqrt{1-x}dx$$

Answer 2

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!