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

Marks : 8

Year : 2013

Auxiliary equation is $$D^3 + 2D^2 +D = 0 \\ D(D^2+ 2D + 1) = 0 \\ D = 0 or (D + 1)^2 = 0 \\ D = 0 or D = -1, -1 \\ Y_c = C_1e^0x + (C_2 + xC_3)e^x $$

Now P.I is $y_p=\dfrac 1{f(D)X} \\ y_p=\dfrac 1{D^3+2D^2+D}(x^2e^{3x}+\sin^2x)+2^x\\ =\dfrac 1{D^3+D^2+D}(x^2e^{3x})+\dfrac 1{D^3+D^2+D}\sin^2x+\dfrac 1{D^3+D^2+D}2^x$

Consider first team

$$ i.e \dfrac 1{D^3+D^2+D}x^2e^{3x}\\ =e^{3x}\dfrac 1{(D+3)^3+2(D+3)^2+(D+3)}x^2 \\ =e^x\Bigg[\dfrac 1{D^3+9D^2+27D+27+2(D^2+6D+9)+D+3}\Bigg]x^2 \\ =e^{3x}\dfrac1{D^3+11D^2+40D+48}x^2 \\ =\dfrac {e^{3x}}{48}\Bigg[1+\Bigg(\dfrac {40D}{48}+\dfrac {11D^2}{48}+\dfrac {D^3}{48}\Bigg)\Bigg]^{-1}x^2 \\ = \dfrac {e^{3x}}{48}\Bigg[1-\Bigg(\dfrac {5D}{6}+\dfrac {11D^2}{48}+\dfrac {D^3}{48}\Bigg)+\Bigg(\dfrac {5D}{6}+\dfrac {11D^2}{48}+\dfrac {D^3}{48}\Bigg)\Bigg]^{-1}x^2 \\ =\dfrac {e^{3x}}{48}\Bigg[1- \dfrac {5D}{6} - \dfrac {11D^2}{48}+\dfrac {25D^2}{36}-\dfrac {D^3}{48}.....\Bigg]x^2 \\ =\dfrac {e^{3x}}{48}\Bigg[x^2-\dfrac 56(2x)-\dfrac {11}{48}\times 2+\dfrac {25}{36}\times 2\Bigg] \\ = \dfrac {e^{3x}}{48}\Bigg[x^2-\dfrac 53x+\dfrac {67}{72}\Bigg] ------- (1)$$

Consider $2^n$ term

$$=\dfrac 1{D^3+2D^2+D}.\sin^2x \\ = \dfrac 1{D^3+2D^2+D}.\dfrac {1-\cos2x}2 \\ =\dfrac 1{D^3+2D^2+D}\times \dfrac 12 e^{0x}- \dfrac 1{D^3+2D^2+D}.\dfrac {\cos 2x}2 \\ =\dfrac 1{2D}\times \dfrac 1{0^2+2(0)+1}e^{0x}-\dfrac 12.\dfrac 1{D(-2)^2+2(-2^2)+D}\cos 2x \\ =\dfrac 1{2D}e^{0x}-\dfrac 12\dfrac 1{-3D-8}\cos 2x \\ =\dfrac x2+\dfrac 12\dfrac 1{3D+8}\times \dfrac {3D-8}{3D-8}\cos2x \\ =\dfrac x2+\dfrac 12\times \dfrac {3D-8}{9D^2-64}\cos 2x \\ =\dfrac x2+\dfrac 12\times \dfrac {3D-8}{9(-2^2)-64}\cos 2x \\ = \dfrac x2+\dfrac 12\times \dfrac {3D\cos2x -8\cos 2x}{-36-64} \\ = \dfrac x2+\dfrac 12\times \dfrac {-3\sin 2x\times 2-8\cos 2x}{-100} \\ =\dfrac x2+\dfrac 12\Bigg(\dfrac {6\sin 2x+8\cos 2x}{100}\Bigg) \\ =\dfrac x2+ \dfrac {3\sin 2x + 4\cos 2x}{100} ------ (2)$$

Considering the third term

$$\dfrac 1{D(D+1)^2}2^x=\dfrac 1{D(D+1)^2}e^{\log 2^{x}} \\ =\dfrac 1{D(D+1)^2}e^{x\log 2} \\ = \dfrac 1 {\log 2(\log 2 +1)^2}2^x ----- (3) $$

Adding (1), (2) and (3) we get

$$y_p=\dfrac {e^{3x}}{48}\Bigg[x^2-\dfrac {5x}3+\dfrac {67}{72}\Bigg]+\dfrac x2+\dfrac 1{100}3\sin 2x + 4\cos 2x+\dfrac 1{\log 2(\log 2+1)^2}2^x$$

The complete solution is

$$y=C_1+(C_2+xC_3)e^x + \dfrac {e^{3x}}{48}\Bigg[x^2-\dfrac {5x}3+\dfrac {67}{72}\Bigg]+\dfrac x2+\dfrac 1{100}3\sin 2x + 4\cos 2x+\dfrac {2^x}{\log 2(\log 2+1)^2}$$