# 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.

• Maple says this here $$y \left( x \right) ={\frac { \left( x+1 \right) {\it \_C1}}{\sqrt {-{x }^{2}+1}}}+2/3\, \left( x-1 \right) \sqrt {x+1}$$ – Dr. Sonnhard Graubner Jan 15 at 18:15

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!
• The condition is $y(0) = 0$ – andrew Jan 15 at 18:52
• yes Robert Z, I obtain that: $$\sqrt{\frac{1-x}{x+1} }\cdot y'(x) + \frac{1}{x^2-1}\cdot\sqrt{\frac{1-x}{x+1} }y = \sqrt{x+1}\cdot\sqrt{\frac{1-x}{x+1} }$$ $$\sqrt{\frac{1-x}{x+1} }*y(x) = \int \sqrt{1-x}dx+c$$ – andrew Jan 16 at 18:30
• then $$y(x) =\Big(\sqrt{\frac{1-x}{x+1}}\Big)^{-1}*\Big[ \int \sqrt{1-x}dx+c\Big]$$ $$y(x) =\Big(\sqrt{\frac{1-x}{x+1}}\Big)^{-1}*\Big[ - \frac{2}{3} \sqrt{(1-x)^3}+c\Big]$$ rembering that $y(0)=0$ well $$y(0)=-\frac{2}{3}+c=0$$ $$c=\frac{2}{3}$$ Then the solution is : $$y(x)=\Big(\frac{1}{\sqrt{\frac{1-x}{x+1}}}\Big)*[ - \frac{2}{3} \sqrt{(1-x)^3}+\frac{2}{3}\Big]$$ – andrew Jan 16 at 18:30
Computing $$\mu(x)=e^{\int\frac{1}{x^2-1}dx}=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ then you will get $$\int\frac{d}{dx}\left(\frac{\sqrt{1-x}y(x)}{\sqrt{x+1}}\right)=\int\sqrt{1-x}dx$$