Question

Find u from the differential equation and initial condition.

du/dt= e^(1.5t-1.3u), u=0 1.3

Find u=?

Answer 1

$\frac{du}{dt} = e^{1.5t-1.3u}$ , $u(0) = 1.3$

$=>\frac{du}{dt}=e^{1.5t}e^{-1.3u}$

$=>e^{1.3u}du=e^{1.5t}dt$

Interacting both sides, we have;

$=>\frac{e^{1.3u}}{1.3}=\frac{e^{1.5t}}{1.5} + c$

Recall that $u(0)=1.3$

$=>\frac{e^{1.3(1.3)}}{1.3}=\frac{e^{1.5(0)}}{1.5} + c$

$=>\frac{e^{1.69}}{1.3}=\frac{e^{0}}{1.5} + c$

$=>4.17=0.67+c$

$=>c=4.17-0.67=3.5$

$=>\frac{e^{1.3u}}{1.3}=\frac{e^{1.5t}}{1.5}+3.5$

Multiply through by 1.3. we have;

$=>e^{1.3u}=\frac{13e^{1.5t}}{15}+\frac{91}{20}$

$=>e^{1.3u}=\frac{52e^{1.5t}+273}{60}$

Take natural log of both sides

$=>lne^{1.3u}=ln(\frac{52e^{1.5t}+273}{60})$

$=>1.3u=ln(\frac{52e^{1.5t}+273}{60})$

$=>u=\frac{10ln(\frac{52e^{1.5t}+273}{60})}{13}$