NIMCET 2019 — Mathematics PYQ

CALCULATION:
Differential equation $f^{\prime }(x)=f(x)$
$f^{\prime }(x)-f(x)=0$
$[D^2-1]f(x)=0$
The corresponding auxiliary equation $D^2-1=0$
Root of auxiliary equation $D^2=1$
⇒ D = ±1
Here, $m=1$ and $n=-1$

Roots are real and different. So complimentary solution is given by
$\Rightarrow f(x)=c_1{e}^x+c_2{e}^{-x}$

Put initial condition f(0)=2 in the above equation we get,
$\Rightarrow 2=c_1{e}^0+c_2{e}^0$
$\Rightarrow 2=c_1+c_2$
-----(1)

$\because f(x)=c_1{e}^x+c_2{e}^{-x}$
$\Rightarrow f(x)=c_1{e}^x-c_2{e}^{-x}$

Put initial condition f(0)=3 in the above equation
$\Rightarrow 3=c_1-c_2$
--------(2)

By adding equation (1) and equation (2) we get,
$\Rightarrow 2c_1=5\Rightarrow c_1=5/2$

Put value of $c_1$ in equation (1)
$\Rightarrow 2=5/2+c_2\Rightarrow c_2=-1/2$

Solution of the given differential equation $f(x)=\frac{5}{2}{e}^x-\frac{1}{2}{e}^{-x}$
$\Rightarrow f(x)=\frac{\left(5e^2-1\right)}{2e^x}$

So, the value of $f(4)$
$\Rightarrow f(4)=\frac{5e^4-1}{2e^4}$