Differential Equations Question 2
Question 2 - 2024 (01 Feb Shift 1)
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
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Answer (14)
Solution
$(t+1) d x=\left(2 x+(t+1)^{4}\right) d t$
$\frac{d x}{d t}=\frac{2 x+(t+1)^{4}}{t+1}$
$\frac{d x}{d t}-\frac{2 x}{t+1}=(t+1)^{3}$
$I \cdot F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^{2}}$
$\frac{x}{(t+1)^{2}}=\int \frac{1}{(t+1)^{2}}(t+1)^{3} d t+c$
$\frac{x}{(t+1)^{2}}=\frac{(t+1)^{2}}{2}+c$
$\Rightarrow c=\frac{3}{2}$
$x=\frac{(t+1)^{4}}{2}+\frac{3}{2}(t+1)^{2}$
put, $t=1$
$x=2^{3}+6=14$
