Answer to Question #191488 in Differential Equations for Ravinder Singh

Question #191488

(3D^2-2D^2+D-1)z=4e^(x+y).cos(x+y)


1
Expert's answer
2021-05-11T11:40:26-0400

Given equation is-

(3D22D2+D1)z=4e(x+y).cos(x+y)(D2+D1)z=4e(x+y).cos(x+y)(3D^2-2D^2+D-1)z=4e^{(x+y)}.cos(x+y)\\[9pt](D^2+D-1)z=4e^{(x+y)}.cos(x+y)


Auxiliary equation is-


m2+m1=0m^2+m-1=0


m=1±142=1±3i2m=\dfrac{-1\pm \sqrt{1-4}}{2}=\dfrac{-1\pm \sqrt{3}i}{2}


Complimentary function is-


CF=ex2(c1cos32x+c2sin32x)CF=e^{-\frac{x}{2}}(c_1cos \dfrac{\sqrt{3}}{2}x+c_2sin\dfrac{\sqrt{3}}{2}x)



Particular integral is-

PI=4ex+ycos(x+y)D2+D1PI=\dfrac{4e^{x+y}cos(x+y)}{D^2+D-1}


=4ex+ycos(x+y)(D+1)2+(D+1)1=4ex+ycos(x+y)D2+3D+1=4ex+ycos(x+y)1+3D+1=4ex+ycos(x+y)3D=4e(x+y)×sin(x+y)3=4e^{x+y}\dfrac{cos(x+y)}{(D+1)^2+(D+1)-1}\\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{D^2+3D+1} \\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{-1+3D+1}\\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{3D}\\[9pt]=-4e^{(x+y)}\times \dfrac{sin(x+y)}{3}


Complete solution is-


y=CF+PI

y=ex2(c1cos32x+c2sin32x)4e(x+y)×sin(x+y)3y = e^{-\frac{x}{2}}(c_1cos \dfrac{\sqrt{3}}{2}x+c_2sin\dfrac{\sqrt{3}}{2}x)-4e^{(x+y)}\times \dfrac{sin(x+y)}{3}



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