Question

Given a function '$g$' whcih has a derivative $g^{\prime }\left(x\right)$ for every real '$x$' and which satisfy $g^{\prime }\left(0\right)=2$ and $g(x+y)=e^y.g\left(x\right)+e^x.g\left(y\right)$ for all $x,y$. Find $g\left(x\right)$.

• A. $2x{e}^x$
• B. $x{e}^x$
• C. $x+e^x$
• D. $x-e^x$

Answer 1

Answer: (A)

$2x{e}^x$

We have

$g^{\prime }\left(0\right)=2$ and

$g(x+y)=e^y.g\left(x\right)+e^x.g\left(y\right)$

Substituting $y=-x$, we get

$g\left(0\right)=e^{-x}.g\left(x\right)+e^x.g(-x)$

If $x=0,g\left(0\right)=2g\left(0\right)\Rightarrow g\left(0\right)=0$ and thus,

$g\left(x\right)+e^{2x}.g(-x)=0$ ... (i)

Now, differentiating the given equation w.r.t. $x$, we have

$g^{\prime }(x+y)=e^y\cdot g^{\prime }\left(x\right)+e^x\cdot g\left(y\right)$... (ii)

Also, differentiating the equation (i) w.r.t. $x$, we have

$g^{\prime }\left(x\right)=e^{2x}\left(g^{\prime }\right(-x)-2g(-x\left)\right)$ ... (iii)

Substituting (iii) in (ii), we get

$g^{\prime }(x+y)=e^y{e}^{2x}\left(g^{\prime }\right(-x)-2g(-x\left)\right)+e^{x}g\left(y\right)$ ...(iv)

Assuming $y=-x$ in (iv), we get

$g^{\prime }\left(0\right)=e^x{g}^{\prime }(-x)-e^{x}g(-x)$

$\Rightarrow 2e^{-x}=g^{\prime }(-x)-g(-x)$

Assuming $-x=X$, we get

$2e^X=g^{\prime }\left(X\right)-g\left(X\right)$

$\Rightarrow g^{\prime }\left(X\right)=g\left(X\right)+2e^X$

Using property of linear differential equation, we get

$g\left(X\right)=\frac{\int 2e^X\cdot e^{\int -1dX}}{e^{\int -1dX}}$

$\Rightarrow g\left(X\right)=2X{e}^X$

Hence, $g\left(x\right)=2x{e}^x$

Constant of integration is $0$ as $g\left(0\right)=0$