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प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
उत्तर
The given equation is \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] ...(1)
where c is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
\[\frac{dy}{dx} = 2\left( 2x \right) + c e^{- x^2} ( - 2x)\]
\[ \Rightarrow \frac{dy}{dx} = 4x - 2xc e^{- x^2} . . . \left( 2 \right)\]
From (1) and (2), we get
\[\Rightarrow \frac{dy}{dx} = 4x - 2xy + 4 x^3 - 4x\]
\[ \Rightarrow \frac{dy}{dx} + 2xy = 4 x^3\]
Hence,
\[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is the solution to the differential equation \[\frac{dy}{dx} + 2xy = 4 x^3\]
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