5(sin(t) (dy)/(dx) +ycos(t)))=cos(t) (sin(t))^(2) for 0\ \textless \ t\ \textless \ pi and y(pi/2) = 9 What is y(t)?

Question

`5(sin(t) (dy)/(dx) +ycos(t)))=cos(t) (sin(t))^(2)`

for
`0\ \textless \ t\ \textless \ pi` and y(pi/2) = 9
What is y(t)?

Answer 1

Assuming you mean


`5\sin t(\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:


`(\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=\frac13\sin^3t+C`

`y=\frac1{15}\sin^2t+C\csc t`

Given that
`y\left(\frac\pi2\right)=9`, we have


`9=\frac1{15}\sin^2\frac\pi2+C\csc\frac\pi2`

`9=\frac1{15}+C`

`C=(134)/(15)`

so that the particular solution over the interval is


`y=\frac1{15}\sin^2t+(134)/(15)\csc t`