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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)?

1 Answer

4 votes
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
User Ujjual
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