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The curve satisfying the differential equation, ydx−(x+3y²)dy=0 and passing through the point (1, 1), also passes through the point :

Option 1)
$\left ( \frac{1}{4} \, , - \frac{1}{2} \right )$

Option 2)
$\left (- \frac{1}{3} ,\: \frac{1}{3} \right )$

Option 3)
$\left (\frac{1}{3} , -\frac{1}{3} \right )$

Option 4)
$\left (\frac{1}{4} ,\frac{1}{2} \right )$

Answers (1)

best_answer

As we learnt in

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

$ydx-\left ( x+3y^{2} \right )dy=0$

$y\frac{dx}{dy}=x+3y^{2}$

$\frac{dx}{dy}-\frac{x}{y}=3y,$ $P=-\frac{1}{y}$ , $Q=3y$

$-\int \frac{1}{y}dy=-logy=log\frac{1}{y}$

$I.F. =e^{log\frac{1}{y}}=\frac{1}{y}$

Solution

$x.\frac{1}{y}=\int \frac{1}{y}\times3ydy =3y$

$3y=\frac{x}{y}+C$

$\therefore 3=1+C$ $\therefore C=2$ $\left [ x=1, y=1 \right ]$

$\therefore 3y=\frac{x}{y}+2$

$y=\frac{1}{3},\: x=\frac{-1}{3}$ Satisfy

$3\times \frac{1}{3} =\frac{\frac{-1}{3}}{\frac{1}{}3}+2$

$1=-1+2 =1$

$\therefore 1=1$

Option 1)

$\left ( \frac{1}{4} \, , - \frac{1}{2} \right )$

Incorrect option

Option 2)

$\left (- \frac{1}{3} ,\: \frac{1}{3} \right )$

Correct option

Option 3)

$\left (\frac{1}{3} , -\frac{1}{3} \right )$

Incorrect option

Option 4)

$\left (\frac{1}{4} ,\frac{1}{2} \right )$

Incorrect option

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