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Solve the initial-value problem.

$ t^3 \frac{du}{dt} + 3t^2y = \cos t, y(\pi) = 0 $

$$

y=\frac{\sin t}{t^{3}}

$$

Differential Equations

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to solve this initial value problem. The first thing they could do is divide each turn by CI Cube to get us into the standard form. Why? Prime plus p of x y is q of axe. So again, I'm dividing each term. Okay, Integrating factor. Each of the integral p of t d t. We know this integrates to eat the three natural log of t which is the same thing as eat the natural log of t cubed. Remember, each the not walk of is simply one which means we have t cubed is our integrating factor multiplying the differential equation by the intrigue factor In other words, each of the terms we end up with us now integrating both sides we can solve for why soul For why, by dividing each of these terms by t cubed order to get why is teach the negative three sign of tea costs each e to the negative three Now use the initial value. Why of pi zero In order find See we have zero is pied to the 93. Just think of this accident. Why, that's literally what we're doing. We're just plugging in our accent. Why values because again we're trying to solve for C, which is our constant of integration. We get zero is equivalent to see because 00 plus c pied to the negative three. So see a zero, which means our solution as why, as a sign of tea divided by t cubed.