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Show that Ax2 + By2 = 1 is a Solution of the Differential Equation X { Y D 2 Y D X 2 + ( D Y D X ) 2 } = Y D Y D X

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प्रश्न

Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]

 

योग
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उत्तर

We have,

\[A x^2 + B y^2 = 1.............(1)\]

Differentiating both sides of (1) with respect to x, we get
\[2Ax + 2By\frac{dy}{dx} = 0 ...........(2)\]
Differentiating both sides of (2) with respect to x, we get
\[2A + 2B \left( \frac{dy}{dx} \right)^2 + 2By\frac{d^2 y}{d x^2} = 0\]
\[ \Rightarrow 2B\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = - 2A\]
\[ \Rightarrow \left[ y\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2 \right] = - \frac{2A}{2B}\]
\[ \Rightarrow \left[ y\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2 \right] = - \left( - \frac{y}{x}\frac{dy}{dx} \right) ...........\left[\text{Using (2)}\right]\]
\[ \Rightarrow x\left[ y\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2 \right] = y\frac{dy}{dx}\]

Hence, the given function is the solution to the given differential equation.

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अध्याय 21: Differential Equations - Exercise 22.03 [पृष्ठ २५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.03 | Q 9 | पृष्ठ २५

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