# Verify that  `y = -tcos(t) - t` is a solution of the initial value problem  `t(dy/dt) = y + t^2sin(t)` ` ` `y(pi) = 0` First we need to find derivative

`dy/dt=-cos t+tsin t-1`

Here we've used product rule to differentiate `-tcos t.`

Now we insert `dy/dt`  and `y` into the initial value problem.

`t(-cos t+tsin t-1)=-tcos t-t+t^2sin t`

`-tcos t+t^2sin t-t=-tcos t-t+t^2sin t`

Obviously, the left and the right sides are equal to each other meaning...

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First we need to find derivative

`dy/dt=-cos t+tsin t-1`

Here we've used product rule to differentiate `-tcos t.`

Now we insert `dy/dt`  and `y` into the initial value problem.

`t(-cos t+tsin t-1)=-tcos t-t+t^2sin t`

`-tcos t+t^2sin t-t=-tcos t-t+t^2sin t`

Obviously, the left and the right sides are equal to each other meaning that the solution satisfies the differential equation. Now we only need to check the initial condition.

`y(pi)=-pi cos pi-pi=-pi(-1)+pi=pi-pi=0`

Therefore, `y=-tcos t-t` satisfies both the differential equation and the initial condition meaning it is the solution to our initial value problem.

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