Jee Main 2024 01 02 2024 Shift 1 - Question6
Question 6
If $(t+1) d x=\left(2 x+(t+1)^{3}\right) d t$ and $x(0)=2$, then $x(1)$ is equal to
(1) 5
(2) 12
(3) 6
(4) 8
Show Answer
Answer: (2)
Solution:
$(t+1) d x=\left(2 x+(t+1)^{3}\right) d t$
$\therefore \frac{d x}{d t}-\frac{2 x}{t+1}=(t+1)^{2}$
$\therefore$ I.F. $=e^{\int-\frac{2}{t+1} d t}=\frac{1}{(t+1)^{2}}$
$\therefore$ Solution is
$\frac{x}{(t+1)^{2}}=\int 1 d t$
$x=(t+c)(t+1)^{2}$ $\because x(0)=2$ then $c=2$
$\therefore x=(t+2)(t+1)^{2}$
$\therefore x(1)=12$
BETA
