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Find the general solution of the differential Equation y'+4y=6cosx

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Solution:   We have ,

                           fracmathrmd^2y mathrmd x^2+4y=6cos x

                   Auxiliary equation is m^2+4=0Rightarrow m=pm 2i

                herefore                C.F=c_1cos 2x+c_2sin 2x

                                  P.I=frac1D^2+4(6cos x)=2cos x

             Now ,        y=C.F+PI=c_1cos 2x+c_2sin 2x+2cos x

Posted by

Deependra Verma

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