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For the Following Differential Equation Verify that the Accompanying Function is a Solution: Differential Equation Function Y = ( D Y D X ) 2 Y = 1 4 ( X ± a ) 2

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प्रश्न

For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[y = \left( \frac{dy}{dx} \right)^2\]
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
योग
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उत्तर

We have,

\[y = \frac{1}{4} \left( x \pm a \right)^2 . . . . . \left( 1 \right)\]

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = \frac{1}{4} \times 2\left( x \pm a \right)\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2}\left( x \pm a \right)\]

Squaring both sides we get

\[ \Rightarrow \left( \frac{dy}{dx} \right)^2 = \left[ \frac{1}{2}\left( x \pm a \right) \right]^2 \]

\[ \Rightarrow \left( \frac{dy}{dx} \right)^2 = \frac{1}{4} \left( x \pm a \right)^2 \]

\[ \Rightarrow \left( \frac{dy}{dx} \right)^2 = y ............\left[\text{Using } \left( 1 \right) \right]\]

\[ \therefore y = \left( \frac{dy}{dx} \right)^2\]

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 21.5 | पृष्ठ २५

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