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A curve passing through \left ( 1,2 \right ) and satisfying \int_{0}^{x}t^{2}y\left ( t \right )dt=x^{3}y\left ( x \right )\left ( x>0 \right ) is

Option: 1

x^{2}y=2


Option: 2

x^{2}+y^{2}=5


Option: 3

\frac{x^{2}}{2}+\frac{y^{2}}{8}=1

 


Option: 4

y^{2}=4x


Answers (1)

best_answer

\int_{0}^{x}t^{2}y\left ( t \right )dt=x^{3}y\left ( x \right )

Now differentiate it w.r.to 'x'

\Rightarrow x^{2}y\left ( x \right )\times 1=x^{3}y'\left ( x \right )+3x^{2}y\left ( x \right )

\Rightarrow x^{3}y'\left ( x \right )=-2x^{2}y\left ( x \right )

\Rightarrow \int \frac{y'\left ( x \right )}{y\left ( x \right )}dx=-\int \frac{2}{x}dx

\Rightarrow In \left ( y\left ( x \right ) \right )=-2 In\; x+c it passes through \left ( 1,2 \right )

So, c=+\; In\; 2

Now put value of

In\left ( y\left ( x \right ) \right )=-2\; In\; x+In^{2}

\Rightarrow yx^{2}=2

Posted by

Sanket Gandhi

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