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

Marks : 6

Year : 2014

$$(D^2+2)y=x^2e^{3x}+e^x-\cos 2x$$

Auxillary equation is $D^2+2=0$

$\therefore D=\pm\sqrt{2i}\\ \therefore C.f \space is\\ y_c=e^{0x}(C_1\cos\sqrt{2x}+C_2\sin \sqrt{2x})$

Now particular integral is $y_p=\dfrac 1{f(D)}x$

$ y_p=\dfrac 1{D+2}[x^2e^{3x}+e^x-\cos 2x] \\ =\dfrac 1{D^2+2}x^2e^{3x}+\dfrac 1{D^2+2}e^x-\dfrac 1{D^2+2}\cos 2x\\ =e^{3x}\dfrac 1{(D+3)^2+2}x^2+e^x\dfrac 1{1^2+2}-\cos 2x\times \dfrac 1{-(2)^2+2}\\ =e^{3x}\dfrac 1{D^2+6D+11}x^2+\dfrac {e^x}3+\dfrac {\cos 2x}2\\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[1+\bigg(\dfrac {D^2+6D}{11}\Bigg)\Bigg]^{-1}x^2\\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[1-\Bigg(\dfrac{D^2+6D}{11}\Bigg)+\Bigg(\dfrac{D^2+6D}{11}\Bigg)\Bigg]x^2 \\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[1- \dfrac {6D}{11}-\dfrac {D^2}{11}+\dfrac {6D^2}{11^2}+...\Bigg]x^2\\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[1-\dfrac {6D}{11}-\dfrac {6D^2}{121}+...\Bigg]x^2 \\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[x^2-\dfrac 6{11}(2x)\dfrac {25}{121}(2)\Bigg]\\ =\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[x^2-\dfrac {12x}{11}+\dfrac {50}{121}\Bigg]$

The complete solution is $y_c+y_p$

$y=C_1\cos\sqrt{2x}+C_2\sin \sqrt{2x}+\dfrac {e^x}3+\dfrac {\cos 2x}2+\dfrac {e^{3x}}{11}\Bigg[x^2-\dfrac {12x}{11}+\dfrac {50}{121}\Bigg]$