Question

t%29%29%5E%7B2%7D%20" id="TexFormula1" title="5(sin(t) \frac{dy}{dx} +ycos(t)))=cos(t) (sin(t))^{2} " alt="5(sin(t) \frac{dy}{dx} +ycos(t)))=cos(t) (sin(t))^{2} " align="absmiddle" class="latex-formula">
for $0\ \textless \ t\ \textless \ pi$ and y(pi/2) = 9
What is y(t)?

Answer 1

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$, and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$, we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$