Let C1 be the curve obtained by the solution of differential equation [2xydy dx = y^2 - x^…

Question

Let C1 be the curve obtained by the solution of differential equation

$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C2 be the

solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :

Accepted Answer

$${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$

Put $y = vx$

$$v + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}$$

$$x{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} =  - {{({v^2} + 1)} \over {2v}}$$

$$ \Rightarrow {{2v} \over {{v^2} + 1}}dv =  - {{dx} \over x}$$

$$\ln ({v^2} + 1) =  - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}$$

$$ \Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx$$

It pass through (1, 1)

$$ \therefore $$ ${x^2} + {y^2} - 2x = 0$

Similarly for second differential equation $${{dy} \over {dx}} = $$ ${{2xy} \over {{x^2} - {y^2}}}$Equation of curve is x2 + y2$-$2y = 0

Now required area is$ = 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}}  - x} \right)} dx $$= ({\pi  \over 2} - 1)$

Solutions · 1