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.

  • $\begingroup$ Maple says this here $$y \left( x \right) ={\frac { \left( x+1 \right) {\it \_C1}}{\sqrt {-{x }^{2}+1}}}+2/3\, \left( x-1 \right) \sqrt {x+1} $$ $\endgroup$ – 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!

  • $\begingroup$ The condition is $y(0) = 0$ $\endgroup$ – andrew Jan 15 at 18:52
  • $\begingroup$ @andrew OK. Have you tried to go on? $\endgroup$ – Robert Z Jan 15 at 18:54
  • $\begingroup$ yes Robert Z, I obtain that: $$ \sqrt{\frac{1-x}{x+1} }\cdot y'(x) + \frac{1}{x^2-1}\cdot\sqrt{\frac{1-x}{x+1} }y = \sqrt{x+1}\cdot\sqrt{\frac{1-x}{x+1} }$$ $$ \sqrt{\frac{1-x}{x+1} }*y(x) = \int \sqrt{1-x}dx+c$$ $\endgroup$ – andrew Jan 16 at 18:30
  • $\begingroup$ 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 : $$y(x)=\Big(\frac{1}{\sqrt{\frac{1-x}{x+1}}}\Big)*[ - \frac{2}{3} \sqrt{(1-x)^3}+\frac{2}{3}\Big]$$ $\endgroup$ – andrew Jan 16 at 18:30
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    $\begingroup$ Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour $\endgroup$ – Robert Z Jan 16 at 18:46

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


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