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)

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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