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Verify that $y=-t \cos t-t$ is a solution of the initial-value problem$$t \frac{d y}{d t}=y+t^{2} \sin t \quad y(\pi)=0$$Is this differential equation pure-time, autonomous, ornonautonomous?

Use the solution to verify if the LHS and RHS of the differential equationequals. For the detailed work, see the solution.

Calculus 2 / BC

Chapter 7

Differential Equations

Section 1

Modeling with Differential Equations

Missouri State University

Harvey Mudd College

Baylor University

Lectures

01:11

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06:22

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02:15

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we can verify that the function wise equals a native T co. Sanity minus. He's a solution to the differential equation as follows. So we substituted function into the left hand side of the differential equation, giving us that the left hand side is equal to T d u I D t. Which is equal to a tee times native co side of T Negative coast, not co tension. That's consigning T minus plus t sign of teeth puts musty sign of teeth side T minus one and we got this use of product rule. If we simple find it down a little bit, we end up that this is the contaminated T close. I have t minus T plus t squared off the sign of teeth, which is equal to why lusty square times, the sine of t, which it is the right inside. So the left hand side is equal to the right hand side. Was implies that this isn't back solution to the differential equation.

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