JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 15)
The curve satifying the differeial equation, (x2 $$-$$ y2) dx + 2xydy = 0 and passing through the point (1, 1) is :
a circle of radius one.
a hyperbola.
an ellipse.
a circle of radius two.
Explanation
(x2 $$-$$ y2) dx + 2xydy = 0
$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{{y^2} - {x^2}} \over {2xy}}$$
Let y = vx
$${{dy} \over {dx}}$$ = v + x $${{dv} \over {dx}}$$
$$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}$$ $$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2} - 1} \over {2v}}$$
$$ \Rightarrow $$ x$${{dv} \over {dx}}$$ = $${{ - {v^2} - 1} \over {2v}}$$
$$ \Rightarrow $$ $${{2vdv} \over {{v^2} + 1}}$$ = $$-$$ $${{dx} \over x}$$
After intergrating, we get
$$\ln \left| {{v^2} + 1} \right|$$ = $$-$$ ln$$\left| x \right|$$ + lnc
$${{y{}^2} \over {{x^2}}}$$ + 1 = $${c \over x}$$
As curve passes through the point (1, 1), so 1 + 1 = c
$$ \Rightarrow $$ c = 2
x2 + y2 $$-$$ 2x = 0, which is a circle of radius one.
$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{{y^2} - {x^2}} \over {2xy}}$$
Let y = vx
$${{dy} \over {dx}}$$ = v + x $${{dv} \over {dx}}$$
$$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}$$ $$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2} - 1} \over {2v}}$$
$$ \Rightarrow $$ x$${{dv} \over {dx}}$$ = $${{ - {v^2} - 1} \over {2v}}$$
$$ \Rightarrow $$ $${{2vdv} \over {{v^2} + 1}}$$ = $$-$$ $${{dx} \over x}$$
After intergrating, we get
$$\ln \left| {{v^2} + 1} \right|$$ = $$-$$ ln$$\left| x \right|$$ + lnc
$${{y{}^2} \over {{x^2}}}$$ + 1 = $${c \over x}$$
As curve passes through the point (1, 1), so 1 + 1 = c
$$ \Rightarrow $$ c = 2
x2 + y2 $$-$$ 2x = 0, which is a circle of radius one.
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