Explanation Given: The given equations are, x ″ + 5 x − 4 y = 0 − x + y ″ + 2 y = 0 Approach: 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. x ″ + 5 x − 4 y = 0 − x + y ″ + 2 y = 0 Write the above linear system of equations in operator form that is, ( D 2 + 5 ) [ x ] − 4 [ y ] = 0 ... ( 1 ) − x + ( D 2 + 2 ) [ y ] = 0 ... ( 2 ) Multiply the equation ( 2 ) with ( D 2 + 5 ) and add the above equations. ( D 2 + 5 ) [ x ] − 4 [ y ] = 0 − ( D 2 + 5 ) x + ( D 4 + 7 D 2 + 10 ) [ y ] = 0 ( D 4 + 7 D 2 + 10 ) [ y ] = 0 … ( 3 ) The auxiliary equation of the above equation is D 4 + 7 D 2 + 10 = 0 D 2 = − 6 , − 1 D = ± i 6 , ± i The general solution of the equation is given by, y = c 1 cos t + c 2 sin t + c 3 cos ( 6 t ) + c 4 sin ( 6 t ) Here, c 1 , c 2 , c 3 and c 4 are arbitrary constants

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ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Differential Equations

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To find:

The general solution for the linear system of equation.