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Show that the Differential Equation of Which Y = 2 ( X 2 − 1 ) + C E − X 2 is a Solution is D Y D X + 2 X Y = 4 X 3

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प्रश्न

Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]

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उत्तर

We have,
\[y = 2\left( x^2 - 1 \right) + c e^{- x^2}...........(1)\]
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dx} = 4x - c e^{- x^2} 2x\]
\[ = 2x\left[ 2 - c e^{- x^2} \right]\]
\[ = - 2x\left[ 2 x^2 - 2 + c e^{- x^2} - 2 x^2 \right]\]
\[ = - 2x\left[ 2\left( x^2 - 1 \right) + c e^{- x^2} - 2 x^2 \right]\]
\[ = - 2x\left[ y - 2 x^2 \right] .............\left[\text{Using }\left( 1 \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = - 2xy + 4 x^3 \]
\[ \Rightarrow \frac{dy}{dx} + 2xy = 4 x^3\]

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 19 | पृष्ठ २५

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