If x = x t is the solution of the differential equation t + 1 d x = 2 x + t + 1 4 d t , x 0 = 2 , then x 1…

Given: $\left(t+1\right)dx=\left(2x+{\left(t+1\right)}^4\right)dt$

$\Rightarrow \frac{dx}{dt}=\frac{2x+{\left(t+1\right)}^4}{\left(t+1\right)}$

$\Rightarrow \frac{dx}{dt}-\frac{2x}{\left(t+1\right)}={\left(t+1\right)}^3$

$\Rightarrow \mathrm{IF}=e^{\int \frac{-2}{t+1}dt}$

$\Rightarrow \mathrm{IF}=e^{-2\log \left|t+1\right|}$

$\Rightarrow \mathrm{IF}=\frac{1}{{\left(t+1\right)}^2}$

Solution of the differential equations is given by,

$\Rightarrow x\times \frac{1}{{\left(t+1\right)}^2}=\int \left(t+1\right)dt$

$\Rightarrow \frac{x}{{\left(t+1\right)}^2}=\frac{t^2}{2}+t+c$

It is given that $x\left(0\right)=2$

$\Rightarrow \frac{2}{{\left(0+1\right)}^2}=\frac{0}{2}+0+c$

$\Rightarrow c=2$

$\Rightarrow \frac{x}{{\left(t+1\right)}^2}=\frac{t^2}{2}+t+2$

Putting $t=1$,

$\Rightarrow \frac{x}{{\left(1+1\right)}^2}=\frac{1^2}{2}+1+2$

$\Rightarrow \frac{x}{4}=\frac{1}{2}+3$

$\Rightarrow x=2+12$

$\Rightarrow x=14$