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) equals

Mathematics · Differential Equations

JEE Main 2024 — 1 February, Shift 1 — Question 21

If $x=x(t)$ is the solution of the differential equation $(t+1) d x=\left(2 x+(t+1)^{4}\right) d t, x(0)=2$ , then, $x(1)$ equals $\qquad$

Numerical answer — enter this value.

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

$\begin{aligned} & \frac{d x}{d t}=\frac{2 x+(t+1)^{4}}{t+1}, & \frac{d x}{d t}-\frac{2 x}{t+1}=(t+1)^{3} \end{aligned}$

$I \cdot F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^{2}}$

$\frac{x}{(t+1)^{2}}=\int \frac{1}{(t+1)^{2}}(t+1)^{3} d t+c$

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

$\Rightarrow \mathrm{c}=\frac{3}{2}$

$x=\frac{(t+1)^{4}}{2}+\frac{3}{2}(t+1)^{2}$

put, $\mathrm{t}=1$

$\mathrm{x}=2^{3}+6=14$

Answer key and solution verified before publishing.

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