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Verify that $ y = -t \cos t - t $ is a solution of the initial value problem.

$ t \frac{ty}{dt} = y + t^{2} \sin t $ $ y = \pi = 0 $

In order to verify that the given solution satisfies the Initial value problem, we need to

1. Verify that the solution satisfies the Initial condition

$y=-t \cos t-t$

The Initial condition is $y(\pi)=0$

Therefore check by substituting $t=\pi$

$y(\pi)=-\pi \cos (\pi)-\pi$

$0=-\pi(-1)-\pi$

$0=\pi-\pi$

Hence verified

2. Verify that the solution satisfies the Differential Equation

$y=-t \cos t-t$

Differentiate

$\frac{d y}{d t}=-\cos t+t \sin t-1$

We have to verify the following differential equation:

$t \frac{d y}{d t}=y+t^{2} \sin t$

Substitute the expressions for $\frac{d y}{d t}$ and $y,$ To get

$t[-\cos t+t \sin t-1]=[-t \cos t-t]+t^{2} \sin t$

$-t \cos t+t^{2} \sin t-t=-t \cos t-t+t^{2} \sin t$

$-t \cos t+t^{2} \sin t-t=-t \cos t-t \pm t^{2} \sin t$

$-t \cos t-t=-t \cos t-t$

Hence verified

Differential Equations

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Mississippi State University

Differential Equations