Explanation Approach: 1. Make the characteristic equation by substituting y ″ = r 2 , y ′ = r . The characteristic equation is, a r 2 + b r + c = 0 Here, the coefficient of y ″ is a , the coefficient of y ′ is b and the coefficient of y is c . 2. Find the roots of the characteristic equation. 3. If roots are real. Case-1 If r = r 1 = r 2 , then the general solution is y ( t ) = c 1 e r t + c 2 t e r t Case-2 If r 1 ≠ r 2 , then the general solution is y ( t ) = c 1 e r 1 t + c 2 e r 2 t 4. If roots are imaginary in the form of r 1 , 2 = α ± β i , then the general solution is y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) Calculation: Consider the given differential equation

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 4, Problem 13RE

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The general solution of the equation 6y″+5y′−4y=0.

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Chapter 4, Problem 12RE

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Chapter 4 Solutions

Introductory Differential Equations

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The general solution of the equation 6 y ″ + 5 y ′ − 4 y = 0 .

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Chapter 4 Solutions