MATH 105 Lecture Notes - Lecture 39: Separation Of Variables

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MATH 105 Lecture Notes - Lecture 38: Inverse Trigonometric Functions

MATH 105 Lecture Notes - Lecture 39: Separation Of Variables

MATH 105 Lecture 40: MATH 105 Lecture 40

Document Summary

A separable differential equation appears in the following form: In order to separate a differential equation, all the y"s need to be multiplied by the derivative and all the x"s must be on the other side of the equation. N(y) dy/dx dx = m(x) dx. Now y is really y(x) so we can use the following substitution: u = y(x) du = y"(x) dx = dy/dx dx. We apply this substitution to the integral to get. Now we can integrate both sides and then substitute back for the u on the left side. Note that this is improper separation of derivatives but we can use it as a baseline to actually arrive at our answer. Solve the following differential equation and determine the interval of validity for the solution. dy/dx = 6y 2 x y(1) = 1/25. Now we have an implicit solution and we apply the initial condition and find the value of c.

Related Questions

Using a technique of integration and the separation-of-variables technique, you'll discover a function whose antiderivative models the shape of an ideal cable suspended from its endpoints. The function depends on a parameter b, which is determined by the properties of the cable, such as the density of the material out of which it is made.

Using principles from physics, it can be shown that when a perfectly flexible, uniformly dense, and inextensible cable is suspended from its endpoints, then it will assume a shape of a curve y = f(x) satisfying the following second-order differential equation:

where Ï is the density of the material of the cable, g is the acceleration due to gravity, and T is the magnitude of the tension in the cable at its lowest point.

We assume our cable is as in the diagram above and y = f(x) is its shape function; thus, not only does y satisfy the "hanging cable differential equation" above, we also have y '(0) = 0. Hence, making the substitution u =dy/dx in our the hanging cable equation, we obtain the first order, initial-value problem satisfied by u:

where b = Ïg/T Solve the preceding initial-value problem, entering your (explicit) solution in the form u = g(x) where g is a function of x in which the constant b is present.

MATH 105 Lecture Notes - Lecture 39: Separation Of Variables

MATH 105 Lecture Notes - Lecture 38: Inverse Trigonometric Functions

MATH 105 Lecture Notes - Lecture 39: Separation Of Variables

MATH 105 Lecture 40: MATH 105 Lecture 40

Document Summary

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Related textbook solutions

Calculus

Single Variable Calculus: Early Transcendentals

CALCULUS:EARLY TRANSCENDENTALS

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