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

Marks : 6

Year : 2014

Auxiliary equation is $D^2+2=0 \\ D^2=-2\\ D=\pm\sqrt{2i}\\ \therefore CF \space is \space Y_c=C_1\cos\sqrt{2x}+C_2\sin \sqrt{2x}\\ P.I\space is\space Y_p=\dfrac 1{f(D)}X\\ Y_p=\dfrac 1{D^2+2}(x^2e^{3x}+e^x\cos x)\\=\dfrac 1{D^2+2}x^2e^{3x}+\dfrac 1{D^2+2}e^x\cos x$

consider first term

$\dfrac 1{D^2+2}x^2e^{3x}=e^{3x}\dfrac 1{(D+3)^2+2}\\ =e^{3x}\Bigg(\dfrac 1{D^2+6D+11}\Bigg)x^2\\ =\dfrac {e^{3x}}{11}\Bigg(\dfrac1{1+\dfrac {D^2+6D}{11}}\Bigg)x^2\\ = \dfrac {e^{3x}}{11}\Bigg(\dfrac1{1+\dfrac {D^2+6D}{11}}\Bigg)^{-1}x^2 \\ =\dfrac {e^{3x}}{11}\Bigg(1-\Bigg(\dfrac {6D+D^2}{11}\Bigg)+ \Bigg(\dfrac {6D+D^2}{11}\Bigg)^2.......\Bigg)x^2 \\ =\dfrac {e^{3x}}{11}\Bigg(1-\dfrac {6D}{11}-\dfrac {D^2}{11}+\dfrac {36D^2}{121}\Bigg)x^2 \\ =\dfrac {e^{3x}}{11}\Bigg(x^2\dfrac {6(2x)}{11}-\dfrac 2{11}-\dfrac {36}{121}x^2\Bigg)\\ =\dfrac {e^{3x}}{11}\Bigg(x^2-\dfrac {12x}{11}+\dfrac {50}{121}\Bigg)....... (1)$

Consider $2^{nd}$ term

$\therefore \dfrac 1{D^2+2}e^x\cos x=e^x\dfrac 1{(D+1)^2+2}\cos x\\ =e^x\dfrac 1{D^2+2D+3}\cos x\\ e^x\dfrac 1{-1^2+2D+3}\cos x\\ =e^x\dfrac 1{2D+2}\cos x\\ =\dfrac {e^x}2\dfrac 1{D+1}\dfrac {D-1}{D-1}\cos x\\ =\dfrac {e^x}2\dfrac {D-1}{D^2-1}\cos x\\ =\dfrac {e^x}2\dfrac {D\cos x -\cos x}{-1^2-1}\\ =\dfrac {e^x}2\dfrac {-\sin x-\cos x}{-2}\\ =\dfrac {e^x}4(\sin x+\cos x)---- (2) $

Adding 1 & 2 we get $Y_p$

$\therefore Y_p=\dfrac {e^3x}{11}\Big(x^2-\dfrac {12}{11}+\dfrac {50}{121}\Big)+\dfrac {e^x}4(\sin x +\cos x)\\ y=y_c+y_p\\ =C_1\cos \sqrt {2x}+C_2\sin \sqrt{2x}+\dfrac {e^{3x}}{11}\Bigg(x^2-\dfrac {12x}{11}+\dfrac {50}{121}\Bigg)+\dfrac {e^{3x}}4(\sin x+\cos x)$