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# Differential equation problem ?

Verify that y = −5t cos(t) − 5t is a solution of the following initial-value problem.

t(dy/dt) = y+ 5t^(2)sin(t) y(pi)=0

y=-5cos(t)-5t ------------->(dy/dt)= ?

LHS: t(dy/dt) = t(?) = 5t^(2)sin(t) − 5t cos(t) − 5t

=? + y = RHS

please help, I don't understand this

### 1 Answer

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- RealProLv 71 month ago
I don't know what trouble you are facing. You put the y(t) you are given into the differential equation you are given. Right side equals left side. y(pi) = 0. The solution is verified.

Assuming you do in fact mean y = −5t cos(t) − 5t and not y = −5 cos(t) − 5t

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