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Numerade Educator

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Problem 17 Medium Difficulty

Solve the initial-value problem.
$ t \frac {du}{dt} = t^2 + 3u, t > 0, u(2) = 4 $

Answer

$$u=-t^{2}+t^{3}$$

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Video Transcript

in order to solve this initial value problem, the first step would be to divide by teeth and arrange our terms into the standard form of the equation. Why prime plus p of X y is Q of acts. If we are to do this and divide each term by t, we end up with this, which we can now read more easily because, like I said, it is in the standard form, Weaken CP of tea is negative. Three over t therefore each of the integral of the pfft. You can see I'm doing this right here. We get eat the negative three natural log of absolute value of tea, which just means it has to be positive, which gives us each the natural log of teach the negative three each. The natural August 1 teach the negative three is the same thing as one divided by t. Cute. Okay, now that we've got this here, we know we can multiply each of the terms than differential equation by this integrating factor in order to give us an equation where we can now better take the right hand into girl again, multiply each term by the integrating factor, integrate both sides solved for you in this context. And then we can plug in our initial value conditions. We get you is negative t squared, plus c t cubed. Okay. Our initial value condition you have to is for Just think of this is accident. Why plug into the subsequent places where they belong again? We're trying to find See? So in the end, we should end up with C equals some value which we got seed to be positive one Which means are you is gonna be negative t squared, plus t cute.