Explanation Given: The given equations are, d x d t + x + d y d t = e 4 t 2 x + d 2 y d t 2 = 0 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 x d t + x + d y d t = e 4 t 2 x + d 2 y d t 2 = 0 Write the above linear system of equations in operator form that is, ( D + 1 ) [ x ] + D [ y ] = e 4 t ... ( 1 ) 2 x + D 2 [ y ] = 0 ... ( 2 ) Multiply the equation ( 1 ) with ( D − 2 ) and subtract the equations. ( D 2 + D ) [ x ] + D 2 [ y ] = 4 e 4 t 2 x + D 2 [ y ] = 0 ( D 2 + D − 2 ) x = 4 e 4 t ... ( 3 ) The auxiliary equation of the above equation is D 2 + D − 2 = 0 ( D + 2 ) ( D − 1 ) = 0 D = − 2 , 1 The general solution of the equation is given by, x = c 1 e − 2 t + c 2 e t Here c 1 and c 2 are arbitrary constants. Assume the particular solution of equation is, x p = λ e 4 t ... ( 4 ) Differentiate above equation with respect to t . x p ′ = 4 λ e 4 t Differentiate above equation again with respect to t . x p ′′ = 16 λ e 4 t Substitute x p for x in equation ( 3 ) . ( D 2 + D − 2 ) x p = 4 e 4 t x p ′′ + x p ′ − 2 x p = 4 e 4 t

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Chapter 5.2, Problem 16E

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To find: The general solution for the linear system of equation.

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To find:

The general solution for the linear system of equation.