Question

Solve $\frac{dy}{dx}=\frac{x\sqrt{(x^2-1)}+y}{\sqrt{(x^2-1)}}$
Given that y=1 when x=1.

• A. $\therefore y[x+\sqrt{(x^2-1)}]=\frac{1}{3}[x^3-{(x^2-1)}^{3/2}]+\frac{2}{3}$ is the solution.
• B. $\therefore y[x-\sqrt{(x^2-1)}]=\frac{1}{3}[x^3-{(x^2+1)}^{3/2}]+\frac{2}{3}$ is the solution.
• C. $\therefore y[x-\sqrt{(x^2-1)}]=\frac{1}{3}[x^3-{(x^2-1)}^{3/2}]-\frac{2}{3}$ is the solution.
• D. None of these.

Answer 1

The correct option is D None of these.

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

Here $P=-\frac{1}{\sqrt{(x^2-1)}}\Rightarrow \int Pdx=\int -\frac{.1}{\sqrt{(x^2-1)}}dx=-\log [x+\sqrt{(x^2-1)}]$

$\therefore I.F.=\frac{1}{x+\sqrt{(x^2-1)}}=\frac{x-\sqrt{(x^2-1)}}{x^2-(x^2-1)}=x-\sqrt{(x^2-1)}$

Multiplying (1) by $I.F.$ we get

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

Integrating both sides, we get

$y.[x-\sqrt{(x^2-1)}]=\int x[x-\sqrt{(x^2-1)}]dx+c=\int [x^2-\frac{1}{2}.2x\sqrt{(x^2-1)}]dx+c$

$=\frac{1}{3}{x}^3-\frac{1}{2}.\frac{2}{3}.{(x^2-1)}^{3/2}+c$

$\Rightarrow y[x-\sqrt{(x^2-1)}]=\frac{1}{3}[x^3-{(x^2-1)}^{3/2}]+c$

Now $x=1,y=1$

$\therefore 1=\frac{1}{3}\left(1\right)+c\Rightarrow c=\frac{2}{3}$

$\therefore y[x-\sqrt{(x^2-1)}]=\frac{1}{3}[x^3-{(x^2-1)}^{3/2}]+\frac{2}{3}$