Explanation Given: The given equations are, d 2 x d t 2 = y ; x ( 0 ) = 3 , x ′ ( 0 ) = 1 d 2 y d t 2 = x ; y ( 0 ) = 1 , y ′ ( 0 ) = − 1 Procedure used: Consider a linear system of two first-order differential equations of the form a 1 x ′ ( t ) + a 2 x ( t ) + a 3 y ′ ( t ) + a 4 y ( t ) = f 1 ( t ) a 5 x ′ ( t ) + a 6 x ( t ) + a 7 y ′ ( t ) + a 8 y ( t ) = f 2 ( t ) Here, a 1 , a 2 , a 3 , ....... , a 8 are constants and x ( t ) and y ( t ) are pair of function to be determined. In operator notation the above two equations become ( a 1 D + a 2 ) [ x ] + ( a 3 D + a 4 ) [ y ] = f 1 ( a 5 D + a 6 ) [ x ] + ( a 7 D + a 8 ) [ y ] = f 2 . Calculation: Consider the given linear system of equations. d 2 x d t 2 = y ; x ( 0 ) = 3 , x ′ ( 0 ) = 1 d 2 y d t 2 = x ; y ( 0 ) = 1 , y ′ ( 0 ) = − 1 Since x = y ″ so, x ′ = y " ' x ″ = y ( 4 ) Write the above linear system of equations in operator form that is, ( D 4 − 1 ) [ y ] = 0 … ( 1 ) The auxiliary equation of the above equation is D 4 − 1 = 0 D 4 = 1 D = 1 , − 1 , − i , i The general solution of the equation is given by, y = c 1 e t + c 2 e − t + c 3 cos t + c 4 sin t … ( 2 ) Here, c 1 , c 2 , c 3 and c 4 are arbitrary constants

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Chapter 5.2, Problem 21E

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The solution for the initial value problem d2xdt2=y;x(0)=3,x′(0)=1 and d2ydt2=x;y(0)=1,y′(0)=−1.

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Chapter 5.2, Problem 20E

Chapter 5.2, Problem 22E

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The solution for the initial value problem d 2 x d t 2 = y ; x ( 0 ) = 3 , x ′ ( 0 ) = 1 and d 2 y d t 2 = x ; y ( 0 ) = 1 , y ′ ( 0 ) = − 1 .

Solution Summary: The author explains the general solution to the given system of equations, namely, lx=et

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To find:

The solution for the initial value problem d2xdt2=y;x(0)=3,x′(0)=1 and d2ydt2=x;y(0)=1,y′(0)=−1.

Want to see the full answer?

Chapter 5 Solutions

The solution for the initial value problem d 2 x d t 2 = y ; x ( 0 ) = 3 , x ′ ( 0 ) = 1 and d 2 y d t 2 = x ; y ( 0 ) = 1 , y ′ ( 0 ) = − 1 .

Solution Summary: The author explains the general solution to the given system of equations, namely, lx=et

Videos

To find: The solution for the initial value problem d2xdt2=y;x(0)=3,x′(0)=1 and d2ydt2=x;y(0)=1,y′(0)=−1.

Want to see the full answer?

Chapter 5 Solutions

To find:

The solution for the initial value problem d2xdt2=y;x(0)=3,x′(0)=1 and d2ydt2=x;y(0)=1,y′(0)=−1.