Explanation The standard form of the higher order differential equation is given below, d n y d x n + a 1 d n − 1 y d x n − 1 + a 2 d n − 2 y d x n − 2 + .................. a n y = f ( x ) If f ( x ) = 0 , then the differential equation is homogeneous. The solution of the homogeneous is equal to complimentary function. If f ( x ) ≠ 0 , then the differential equation is non- homogeneous. The expression for corresponding complimentary function is given below, y = e m x ( c 1 + c 2 t ) …… ( 1 ) The expression for the solution of the quadratic equation is given below, x = − b ± b 2 − 4 a c 2 a The solution of the non-homogeneous is equal to C . F + P . I , where C . F is complementary function and P . I is particular integral. Calculation: Consider the given initial value problem. y ″ − 2 y ′ + y = 0 Consider the auxiliary equation of the above differential equation. D 2 − 2 D + 1 = 0 Here, b = − 2 , a = 1 and c = 1 . Further solve the above equation to find the value of D , D = 2 ± 4 − 4 2 D = 1 Here, m = 1 . Substitute 1 for m in the equation ( 1 ) . y = e t ( c 1 + c 2 t ) …… ( 2 ) Differentiate the above equation.

Fundamentals of Differential Equations and Boundary Value Problems

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ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.3, Problem 26E

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The solution of the initial value problem y″−2y′+y=0 with initial condition y(0)=1 and y′(0)=−2.

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Chapter 4.3, Problem 25E

Chapter 4.3, Problem 27E

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Chapter 4 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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The solution of the initial value problem y ″ − 2 y ′ + y = 0 with initial condition y ( 0 ) = 1 and y ′ ( 0 ) = − 2 .

Solution Summary: The author explains that the solution of the initial value problem y′′-2

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To find:

The solution of the initial value problem y″−2y′+y=0 with initial condition y(0)=1 and y′(0)=−2.

Want to see the full answer?

Chapter 4 Solutions

The solution of the initial value problem y ″ − 2 y ′ + y = 0 with initial condition y ( 0 ) = 1 and y ′ ( 0 ) = − 2 .

Solution Summary: The author explains that the solution of the initial value problem y′′-2

Videos

To find: The solution of the initial value problem y″−2y′+y=0 with initial condition y(0)=1 and y′(0)=−2.

Want to see the full answer?

Chapter 4 Solutions

To find:

The solution of the initial value problem y″−2y′+y=0 with initial condition y(0)=1 and y′(0)=−2.