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Verify that Y2 = 4ax is a Solution of the Differential Equation Y = X D Y D X + a D X D Y

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प्रश्न

Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]

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उत्तर

We have, \[y^2 = 4ax ...........(1)\]

Differentiating both sides of (1) with respect to x, we get
\[2y\frac{dy}{dx} = 4a\]
⇒ \[\frac{dy}{dx} = \frac{2a}{y}  ...........(2)\]

Now, differentiating both sides of (1) with respect to y, we get
\[2y = 4a\frac{dx}{dy}\]
⇒ \[\frac{dx}{dy} = \frac{y}{2a}..............(3)\]

\[\therefore x\frac{dy}{dx} + a\frac{dx}{dy} = x\left( \frac{2a}{y} \right) + a\left( \frac{y}{2a} \right) ..........\left[\text{Using  (2) and (3)}\right]\]
\[ \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{2ax}{y} + \frac{y}{2}\]
\[ \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{y^2}{2y} + \frac{y}{2} ..........\left[\text{Using (1)}\right]\]
\[ \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{y}{2} + \frac{y}{2}\]
\[ \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = y\]
\[\Rightarrow y = x\frac{dy}{dx} + a\frac{dx}{dy}\]

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 8 | पृष्ठ २५

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