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  • Please,please help me A curve passes through the point (0,1) and the gradient at (x,y) on it is y(xy-1). The equation of the curve is

A curve passes through the point (0,1) and the gradient at (x,y) on it is y(xy-1). The equation of the curve is

  • Option 1)

    y(x-1)=1

  • Option 2)

    y(x+1)=1

  • Option 3)

    x(y+1)=1

  • Option 4)

    x(y-1)=1

 

Answers (1)

As we learnt in 

Bernoulli's Equation -

\frac{dy}{dx}+py =Qy^{n}

- wherein

P,Q are the function of x alone.

 

 \frac{dy}{dx}=y(xy-1)

\frac{1}{y^{2}}\frac{dy}{dx}+\frac{1}{y}=x

Let      \frac{1}{y}=t\:\:\:\therefore\:-\frac{1}{y^{2}}\frac{dy}{dx}=\frac{dt}{dx}

\therefore\:-\frac{dt}{dx}+t=x

\therefore\: \frac{dt}{dx}-t=\:-x\:\:\:\:\:\:\:[\because\:\frac{dy}{dx}+Py=\theta form]

\therefore\:sol^{n}\:is\:t.e^{-x}=\int -\:xe^{-x}

\therefore\:\frac{1}{y}\:e^{-x}=e^{-x}x+e^{-x}+c

Put x=0, y=1, c=0

\therefore\:1=y(x+1)

 


Option 1)

y(x-1)=1

Incorrect

Option 2)

y(x+1)=1

Correct

Option 3)

x(y+1)=1

Incorrect

Option 4)

x(y-1)=1

Incorrect

Posted by

Vakul

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