Question

Solve the differential equation: $(x^2-1)\frac{dy}{dx}+2(x+2)y=2(x+1).$

• A. $\frac{y{(x+1)}^3}{x+1}=\{\frac{{(x+1)}^2}{2}-4(x+1)+4\log (x+1)\}+c.$
• B. $\frac{y{(x-1)}^3}{x+1}=\{\frac{{(x+1)}^2}{2}-4(x+1)+4\log (x+1)\}+c.$
• C. $\frac{y{(x-1)}^3}{x-1}=\{\frac{{(x+1)}^2}{2}-4(x+1)+4\log (x+1)\}+c.$
• D. None of these.

Answer 1

The correct option is B $\frac{y{(x-1)}^3}{x+1}=\{\frac{{(x+1)}^2}{2}-4(x+1)+4\log (x+1)\}+c.$

Given, $(x^2-1)\frac{dy}{dx}+2(x+2)y=2(x+1)$
$\Rightarrow \frac{dy}{dx}+\frac{2(x+2)}{x^2-1}y=\frac{2(x+1)}{x^2-1}=\frac{2}{(x-1)}$ ...(1)
Here $P=\frac{2(x+2)}{x^2-1}\Rightarrow \int Pdx=\int \frac{2(x+1)}{x^2-1}dx.$
$=3\log (1-x)-\log (x+1)=\log \left(\frac{{(1-x)}^3}{x+1}\right)$
$\therefore I.F=e^{\log \left(\frac{{(1-x)}^3}{x+1}\right)}=\left(\frac{{(1-x)}^3}{1+x}\right)$
Multiplying (1) by $I.F,$ we get
$\frac{{(1-x)}^3}{1+x}\frac{dy}{dx}-\frac{2{(1-x)}^2(x+2)}{(x+1)}y=-\frac{2{(1-x)}^2}{1+x}$
Integrating both sides
$y\frac{{(x-1)}^3}{1+x}=-\int \frac{2{(1-x)}^3}{1+x}dx+c$
$=2\int \frac{{(x-1)}^3}{x+1}+c=2\int (x^2-4x-\frac{8}{x+1}+7)dx+c$
$=\frac{{(x+1)}^2}{2}-4(x+1)+4\log (x+1)+c$