JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 5)
The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :
a circle with centre on the y-axis
an ellipse with major axis along the y-axis
a circle with centre on the x-axis
a hyperbola with transverse axis along the x-axis
Explanation
(x2 $$-$$ y2) dx + 2xy dy = 0
$${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$
Put $$y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
Solving we get,
$$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } $$
ln(v2 + 1) = $$-$$ ln x + C
(y2 + x2) = Cx
1 + 1 = C $$ \Rightarrow $$ C = 2
y2 + x2 = 2x
$${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$
Put $$y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
Solving we get,
$$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } $$
ln(v2 + 1) = $$-$$ ln x + C
(y2 + x2) = Cx
1 + 1 = C $$ \Rightarrow $$ C = 2
y2 + x2 = 2x
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