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Solve the initial-value problem.$ t \frac {du}{dt} = t^2 + 3u, t > 0, u(2) = 4 $
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Calculus 2 / BC
Chapter 9
Differential Equations
Section 5
Linear Equations
Missouri State University
Campbell University
Oregon State University
University of Nottingham
Lectures
13:37
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.
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Solve the initial-value pr…
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Solve the given initial-va…
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Solve the initial value pr…
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.
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