Answer
$x=-(4-t)^2+5(4-t)$
Work Step by Step
We are given:
$\frac{dx}{dt}+\frac{2}{4-t}x=5$
Integrating factor:
$I(t)=e^{\int \frac{2}{4-t}dt}=e^{-2\ln (-4t)}=\frac{1}{(4-t^2)}$
Multiply both sides by the intergrating factor:
$\frac{1}{(4-t^2)}\frac{dx}{dt}+\frac{1}{(4-t^2)}\frac{2}{4-t}x=5\frac{1}{(4-t^2)}$
$\frac{d}{dt}(\frac{x}{(4-t)^2})=\frac{5}{(4-t)^2}$
Integrate both sides:
$(\frac{x}{(4-t)^2})=\int \frac{5}{(4-t)^2}=\frac{5}{4-t}+c$
The general solution is:
$x=c(4-t)^2+5(4-t)$
Since $x(0)=4$
$4=c(4-0)^2+5(4-0)$
Solve for $c$:
$c=-1$
Hence the particular solution is $x=-(4-t)^2+5(4-t)$
