Verify that `y= -t*cos t - t`  is a solution of the initial-value problem `t*(dy/dt)=y+(t^2)sint` and `y(pi) = 0`

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Given that `y = -t*cos(t)-t`

`dy/dt = -t*(-sin t) - cos t - 1`

= `t*sin t - cos t - 1` ...(1)

`y/t + t*sin t `

= `(-t*cos(t)-t)/t + t*sin t`

= `(-cos(t)-1) + t*sin t`

=> `t*sin t - cos t - 1` ...(2)

As (1) =...

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Given that `y = -t*cos(t)-t`

`dy/dt = -t*(-sin t) - cos t - 1`

= `t*sin t - cos t - 1` ...(1)

`y/t + t*sin t `

= `(-t*cos(t)-t)/t + t*sin t`

= `(-cos(t)-1) + t*sin t`

=> `t*sin t - cos t - 1` ...(2)

As (1) = (2)

`t*(dy/dt) = y + t^2*sin t`

`y(pi) = -pi*cos pi - pi`

= `pi - pi`

= 0

This proves that `t*(dy/dt) = y + t^2*sin t` and `y(pi) = 0` .

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