Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2015

Auxillary equation is $D^2 - 4D+3=0\\ D = 3, 1\\ \therefore y_c=C_1e^{3x}+C_2e^x\\ PI ,i.e., y_p=\dfrac 1{f(D)}X\\ \therefore y_p=\dfrac 1{(D-3)(D-1)}(2xe^{3x}+3e^x\cos 2x)\\ y_p=\dfrac 1{(D-3)(D-1)}2xe^{3x}+ \dfrac 1{(D-3)(D-1)}3e^x\cos 2x\\ =2e^{3x}\dfrac 1{(D-3+3)}x+3e^x\dfrac 1{(D+1-3)(D+1-1)}(\cos 2x)\\ =\dfrac {2e^{3x}(1)}{(D+2)D}x+3e^x\dfrac 1{(D-2)}.\dfrac 1D\cos 2x\\ =2e^{2x}\dfrac 1{2\Big(1+\dfrac D2\Big)}\times \dfrac 1Dx+3e^x\dfrac 1{D^2-2D}\cos 2x\\ =e^{3x}\dfrac 1D\Big(1+\dfrac D2\Big)^{-1}x+3e^x\dfrac 1{-2^2-2D}\cos 2x\\ = e^{3x}\dfrac 1D\Big(1-\dfrac D2+\dfrac {D^2}4....\Big)x+ \dfrac {3e^x}{-2}\dfrac 1{D+2}\dfrac {D-2}{D-2}\cos 2x\\ =e^{3x}\dfrac 1D\Big(x-\dfrac 12+0\Big) \dfrac {-3e^x}2\dfrac {(D-2)}{D^2-4}\cos 2x\\ =e^{3x}\int \Big(x-\dfrac 12\Big)dx -\dfrac {-3}2e^x\dfrac {D\cos 2x-2\cos 2x}{-2^2-4}\\ =e^x\Big(\dfrac {x^2}2-\dfrac x2\Big) +\dfrac 3{16}e^x (2(-\sin 2x)-2\cos 2x)\\ =\dfrac {e^{2x}}2x(x-1)-\dfrac 38e^x(\sin 2x+\cos 2x)$