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Solve the differential equation.$ \frac {du}{dt} = \frac {1 + t^4}{ut^2 + u^4t^2} $

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$\frac{u^{2}}{2}+\frac{u^{5}}{5}=-\frac{1}{t}+\frac{t^{3}}{3}+C$

01:10

Amrita Bhasin

01:51

Bobby Barnes

Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

Campbell University

Oregon State University

Boston College

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 differential equ…

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Solve the given differenti…

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Solve the following differ…

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to solve this referential equation. Not that we can erect this into D. Over DT. This is equal to one plus T. To the forest power over T squared times U. Plus U. To the fourth power. Now this is a separate double differential equation. And we can write this as U. Plus U. to the 4th power times D. You. This is equal to one plus T. To the fourth power over T squared DT. Which is the same as You. Bless You. To the 4th power do you? This is equal to to erase the negative too plus T squared. And then DT. And then integrating both sides. We have U. Squared over two plus. Used to the 5th power over five. This is equal to T raised the negative two plus one. That's negative one Over negative one plus. Due to the 3rd power over three and then plus C. Simplifying. We get one half U. Squared Plus 1/5 uses the 5th power. This is equal to negative one over T. Plus one. Third. Keep to the 3rd Power Plus C. Multiplying this by the L. C. D. Which is 30. We have 15 U. Squared plus six U. to the 5th. This is equal to -30 over tea plus 10 T. Race to the 3rd Power Plus C. And so this is the solution to the differential equation.

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