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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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Missouri State University

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University of Nottingham

first, I want to verify that this has a initial value of while pipe will zero. So we're just gonna plug in pie for every t value. So negative pine times coast on a pie minus pi. I don't want this equal zero. So cousin pie is the quadrants of angle pies. Located here does zero. I have high and three pie, perhaps, and pies Here this coordinate is negative 10 and that's co sign sign. So comes on in these cases, negative one. Still have zero equals negative pi times negative one because co Santa pies negative points all minus pi. So zero equals in a negative times negative gives us a positive pot and anything might it still is zero. So this initial condition checks out, we also need to verify that it is so It is a solution of this equation here. So what we have to do is we have to play in our derivative and we have to plug in our original Why equation? So our suspect, the derivative of our y equations. So we have d y B c equals the first times the derivative of the second, which is negative, scientific, close the second times, the derivative of the first, which is negative point. And then the river of, uh, minus T that's just minus one. So we simplify. This will have the positive t Sorrenti minus a co sign T minus one. Just most applied everything together and we simplified. So this is far as we can simplify the situation. Now we need to plug in our were C Y d T equation in our Y collision. So have I t times t sai Inti minus co sign T minus one. Now, when this starts equals, we're gonna play guitar y equation. Our original migration here. So we had negative t co sign t minus t and we bring down our plus t squared sine ti. So now we distribute owner left side will have t squared sign T minus t co sci fi minus E on left side, on the right side. We don't have anything to distribute, So we're just gonna bring everything down. If you look on the left hand side, it matches our right hand side. We have everything. We have a t squared Zion T we have a minus. Seiko Cy Inti and we have a minus t. So since our looks inside, it was our right hand side. We have verified that these solutions

Mississippi State University

Differential Equations