Question

STATEMENT -1 : Solution of $(1+x\sqrt{x^2+y^2})dx+y(-1+\sqrt{x^2+y^2})dy=0$ is $x-\frac{y^2}{2}+\frac{1}{3}(x^2+y^2{)}^{3/2}+C_1=0,C_1$ being arbitrary constant.
STATEMENT-2 : Solution of $xdy-ydx=\sqrt{x^2-y^2}dx$ is $si{n}^{-1}\left(\frac{y}{x}\right)=lnx+C_2,C_2$ being arbitrary constant.

• A. STATEMENT-1 is True, STATEMENT-2 is True ;STATEMENT-2 is a correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True ;STATEMENT-2 is NOT a correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

Answer: (B)

STATEMENT-1 is True, STATEMENT-2 is True ;STATEMENT-2 is NOT a correct explanation for STATEMENT-1

Statement-1 :

$(1+x\sqrt{x^2+y^2})dx+y(-1+\sqrt{x^2+y^2})dy=0$

i.e. $dx-ydy+\frac{1}{2}\sqrt{x^2+y^2}(2xdx+2ydy)=0$

$\int dx-\int ydx+\frac{1}{2}\int \sqrt{x^2+y^2}(2xdx+2ydy)dx=0$

Put $x^2+y^2=t$

$\Rightarrow 2x+2ydy=dt$

So,$x-\frac{y^2}{2}+\frac{1}{2}\int \sqrt{t}dt=0$

$x-\frac{y^2}{2}+\frac{1}{3}{t}^{3/2}+c=0$

$\therefore$ solution is $x-\frac{y^2}{2}+\frac{1}{3}(x^2+y^2{)}^{3/2}+c=0$

$\therefore$ Statement (i) is true.

Statement-2 : $xdy-ydx=\sqrt{x^2-y^2}dx$

$\Rightarrow \frac{xdy-ydx}{x^2}=\sqrt{1-{\left(\frac{y}{x}\right)}^2}\frac{dx}{x}$

$\Rightarrow \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}$

$\Rightarrow si{n}^{-1}\left(\frac{y}{x}\right)=lnx+c$

Statement is true.

Statement 1 is not explained by statement 2. Hence option B is correct.