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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 33 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 33 maths textbook solution

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Answer: x^{2}-y^{2}=-a^{2}

Given: The product of the slope and the ordinate is equal to the abscissa i.e. y(slope of tangent)

= x

To find: The equation of the curves that pass through the point \left ( 0,a \right ).

Hint: Use  y \times \text { slope of tangent }=x to solve.

Solution: we have y(slope of tangent) = x

        \begin{aligned} &=y \frac{d y}{d x}=x \\\\ &=y d y=x d x \end{aligned}

Integrating on both sides we get

        \begin{aligned} &=\int y d y=\int x d x \\\\ &=\frac{y^{2}}{2}=\frac{x^{2}}{2}+C \ldots(i) \end{aligned}

The curve passes through \left ( 0,a \right )

        \begin{aligned} &=\frac{a^{2}}{2}=\frac{0^{2}}{2}+C \\\\ &=\frac{a^{2}}{2}=C \\\\ &=C=\frac{a^{2}}{2} \end{aligned}

Substituting in equation (i) we get

        =\frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{a^{2}}{2}

Multiply 2 on both sides

        \begin{aligned} &=y^{2}=x^{2}+a^{2} \\\\ &=x^{2}-y^{2}=a^{2} \end{aligned}

Hence, required equation of curve is found.

 

 

Posted by

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