Question

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

• A. $a^2\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=e^x$
• B. $a^2\frac{d^{2}y}{d{x}^2}+y=e^x$
• C. $a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=0$
• D. $a\frac{d^{2}y}{d{x}^2}+y=e^x$

Answer 1

The correct option is C $a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=0$

Given, Differential equation as $a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=e^x$ has $f\left(x\right),g\left(x\right)$ as two solutions.

On substituting $f\left(x\right)$ in the given Differential equation,

$\Rightarrow a{f}^{\prime \prime }\left(x\right)+x^2{f}^{\prime }\left(x\right)+f\left(x\right)=e^x................A$

On substituting $g\left(x\right)$ in the given differential equation,

$\Rightarrow a{g}^{\prime \prime }\left(x\right)+x^2{g}^{\prime }\left(x\right)+g\left(x\right)=e^x................B$

Subtracting B from A,

$\Rightarrow a\times \left(f^{\prime \prime }\right(x)-g^{\prime \prime }(x\left)\right)+x^2\times \left(f^{\prime }\right(x)-g^{\prime }(x\left)\right)+\left(f\right(x)-g(x\left)\right)=0$

It is in the form of $a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=0$, which have a solution of $y=f\left(x\right)-g\left(x\right)$