If ⨍ (x) and g(x) are two solutions of the differential equation ad2ydx2+x2dydx+y=ex, then ⨍ (x)-g(x) is the solution of

Given Equation a d2ydx2+x2dydx+y=exf(x) and g(x) are solutions⇒ ad2(f(x))dx2+x2d(f(x))dx+f(x)=ex−−(1)and a d2(g(x))dx2+x2d(g(x))dx+g(x)=ex−−(2)(1)−(2)⇒ ad2dx2(f(x)−g(x))+x2ddx(f(x)−g(x))+f(x)−g(x)=0∴ f(x)−g(x) is solution of a d2ydx2+x2dydx+y=0

If ⨍ (x) and g(x) are two solutions of the differential equation ad2ydx2+x2dydx+y=ex, then ⨍ (x)-g(x) is the solution of

Given Equation a d2ydx2+x2dydx+y=exf(x) and g(x) are solutions⇒ ad2(f(x))dx2+x2d(f(x))dx+f(x)=ex−−(1)and a d2(g(x))dx2+x2d(g(x))dx+g(x)=ex−−(2)(1)−(2)⇒ ad2dx2(f(x)−g(x))+x2ddx(f(x)−g(x))+f(x)−g(x)=0∴ f(x)−g(x) is solution of a d2ydx2+x2dydx+y=0

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If $⨍ (x)$ and $g (x)$ are two solutions of the differential equation $a \frac{d^{2} y}{{dx}^{2}} + x^{2} \frac{dy}{dx} + y = e^{x},$ then $⨍ (x)-g(x)$ is the solution of

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$a^{2} \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} + y = e^{x}$

$a \frac{d^{2} y}{{dx}^{2}} + y = e^{x}$

$a \frac{d^{2} y}{{dx}^{2}} + x^{2} \frac{dy}{dx} + y = 0$

$a^{2} \frac{d^{2} y}{{dx}^{2}} + y = e^{x}$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n a \frac{d^{2} y}{d x^{2}} + x^{2} \frac{d y}{d x} + y = e^{x} \\ f (x) a n d g (x) a r e s o l u t i o n s \\ \Rightarrow a \frac{d^{2} (f (x))}{d x^{2}} + x^{2} \frac{d (f (x))}{d x} + f (x) = e^{x} - - (1) \\ a n d a \frac{d^{2} (g (x))}{d x^{2}} + x^{2} \frac{d (g (x))}{d x} + g (x) = e^{x} - - (2) \\ (1) - (2) \\ \Rightarrow a \frac{d^{2}}{d x^{2}} (f (x) - g (x)) + x^{2} \frac{d}{d x} (f (x) - g (x)) + f (x) - g (x) = 0 \\ ∴ f (x) - g (x) i s s o l u t i o n o f a \frac{d^{2} y}{d x^{2}} + x^{2} \frac{d y}{d x} + y = 0 \end{array}$