Solve using the method of undetermined coefficients y 4y e

Solution

Solve using the method of undetermined coefficients: y\'\' + 4y = e^(3x)cosh(2x) remember that cosh(2x) = (e^(2x) + e^(-2x) )/2 y\'\' + 4y = e^(3x) ( e^(2x) + e^(-2x) ) /2 y\'\' + 4y = e^(5x) / 2 + e^(x) / 2 The root of the characteristic equation r^2+4=0 are r=-2i and r=2i. homogeneous solution: yh= C1 e(-2ix) + C2 e(2ix) = C1( cos(2x) - i sin(2x) ) + C2( cos(2x) + i sin(2x) ) yh= (C1+C2) cos (2x) + (-i C1 + iC2) sin (2x) yh= A cos(2x) + B sin(2x) where A= C1+C2 AND B= (-i C1 + iC2) particular solution is: yp1 = k e^(5x) y\'= 5k e^(5x) y\'\'= 25k e^(5x) y\'\' + 4y = e^(5x) / 2 25k e^(5x) + 4( k e^(5x) ) = e^(5x) / 2 29 ke^(5x) = e^(5x) / 2 29 k = 1/2 k= 1/58 yp1= (1/58) e^(5x) yp2 = k e^(x) y\'= k e^(x) y\'\'= k e^(x) y\'\' + 4y = e^(x) / 2 k e^(x) + 4( k e^(x) ) = e^(x) / 2 5 k e^(x) = e^(x) / 2 5 k = 1/2 k= 1/10 yp2= (1/10) e^(x) yp= yp1+yp2 = (1/58) e^(5x) + (1/10) e^(x) genereal solution is y(x) = yh + yp= A cos(2x) + B sin(2x) + (1/58) e^(5x) + (1/10) e^(x)