Question

$$(x^2-1)\frac{dy}{dx}+2(x+2)y=2(x+1)$$

Answer 1

We have,
$$(x^2-1)\frac{dy}{dx}+2(x+2)y=2(x+1)$$
$$\Rightarrow \frac{dy}{dx}+\frac{2(x+2)}{x^2-1}y=\frac{2(x+1)}{x^2-1}$$
$$\Rightarrow \frac{dy}{dx}+\frac{2(x+2)}{x^2-1}y=\frac{2}{x-1}............\left(1\right)$$
Clearly, it is a linear differential equation of the form 
$$\frac{dy}{dx}+Py=Q$$
where
$$P=\frac{2(x+2)}{x^2-1}$$
$$Q=\frac{2}{x-1}$$
$$\therefore I.F.=e^{\int Pdx}$$
$$=e^{\int \frac{2(x+2)}{x^2-1}dx}$$
$$=e^{\int \frac{2x}{x^2-1}+4\int \frac{1}{x^2-1}dx}$$
$$=e^{log|x^2-1|+4\times \frac{1}{2}\log \left|\frac{x-1}{x+1}\right|}$$
$$=e^{log|(x^2-1)\times \frac{{(x-1)}^2}{{(x+1)}^2}|}$$
$$=e^{log\left|\frac{{(x-1)}^3}{(x+1)}\right|}$$
$$=\frac{{(x-1)}^3}{(x+1)}$$
$$\text{ Multiplying both sides of }\left(1\right)\text{ by }\frac{{(x-1)}^3}{(x+1)},\text{ we get }$$
$$\frac{{(x-1)}^3}{(x+1)}(\frac{dy}{dx}+\frac{2(x+2)}{x^2-1}y)=\frac{{(x-1)}^3}{(x+1)}\times \frac{2}{x-1}$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}\frac{dy}{dx}-\frac{2(x+2){(x-1)}^2}{{(x+1)}^2}y=\frac{2{(x-1)}^2}{(x+1)}$$
Integrating both sides with respect to x, we get
$$\frac{{(x-1)}^3}{(x+1)}y=\int \frac{2{(x-1)}^2}{(x+1)}dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=\int 2\left\{\frac{{(x+1)}^2-4x}{(x+1)}\right\}dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=\int 2\{(x+1)-\frac{4x}{(x+1)}\}dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=\int 2\{(x+1)-\frac{4(x+1-1)}{(x+1)}\}dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=\int 2\{(x+1)-4+\frac{4}{(x+1)}\}dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=\int (2x-6+\frac{8}{x+1})dx+C$$
$$\Rightarrow \frac{{(x-1)}^3}{(x+1)}y=x^2-6x+8\log |x+1|+C$$
$$\Rightarrow y=\frac{(x+1)}{{(x-1)}^3}(x^2-6x+8\log |x+1|+C)$$
$$\text{Hence, }y=\frac{(x+1)}{{(x-1)}^3}(x^2-6x+8\log |x+1|+C)\text{ is the required solution.}$$