The curve satisfying the differential equation- x 2- y 2 dx -2 xy dy -0 and passing through the point 1-1 is -A- a circle of radius twoB- an ellipseC- a hyperbolaD- a circle of radius one

(x^2 y^2) dx+2xy dy=0 ⇒dydx=12 yx(1 ( xy)^2) Put yx=v⇒dydx=v+xdvdx ⇒ v+xdvdx=v2(1 1v^2) ⇒2vv^2+1dv= dxx Integrating both side ⇒∫2vv^2+1dv= ∫dxx ⇒ln (v^2+1)= ln ...

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Standard XII

Mathematics

Homogeneous Differential Equation

The curve sat...

Question

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point $(1, 1)$ is :

a circle of radius one

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a circle of radius two

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an ellipse

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a hyperbola

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Solution

The correct option is A a circle of radius one
$(x^{2} - y^{2}) d x + 2 x y d y = 0$
$\Rightarrow \frac{d y}{d x} = \frac{1}{2} \frac{y}{x} (1 - {(\frac{x}{y})}^{2})$
Put $\frac{y}{x} = v \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$\Rightarrow v + x \frac{d v}{d x} = \frac{v}{2} (1 - \frac{1}{v^{2}})$
$\Rightarrow \frac{2 v}{v^{2} + 1} d v = - \frac{d x}{x}$
Integrating both side
$\Rightarrow \int \frac{2 v}{v^{2} + 1} d v = - \int \frac{d x}{x}$
$\Rightarrow ln (v^{2} + 1) = - ln x + ln c$
$\Rightarrow \frac{y^{2} + x^{2}}{x^{2}} = \frac{c}{x}$
$\Rightarrow c = \frac{y^{2} + x^{2}}{x}$
At $(1, 1), c = 2$
$\Rightarrow x^{2} + y^{2} - 2 x = 0$
$∴$ radius $r = 1$

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The curve satisfying the differential equation, (x2−y2) dx+2xy dy=0 and passing through the point (1,1) is :

The correct option is A a circle of radius one(x2−y2) dx+2xy dy=0 ⇒dydx=12yx(1−(xy)2) Put yx=v⇒dydx=v+xdvdx ⇒v+xdvdx=v2(1−1v2) ⇒2vv2+1dv=−dxx Integrating both side ⇒∫2vv2+1dv=−∫dxx ⇒ln(v2+1)=−lnx+lnc ⇒y2+x2x2=cx ⇒c=y2+x2x At (1,1), c=2 ⇒x2+y2−2x=0 ∴ radius r=1

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point $(1, 1)$ is :