(D2-4D+4)y=e2x+cos2x
Answers
Answer:
Step-by-step explanation:
auxiliary eqn. is m2–4m+4=0⟹(m−2)2=0
m−2,2
Complimentary function
yc=(C1x+C2)e2x
Particular Integral
yp=1(D−2)2x2e2xcos2x
=R.P. of 1(D−2)2x2e(2+i)x
=R.P. of e(2+2i)x1(D+2i)2x2
=R.P. of e(2+2i)x1D2−4+i2Dx2
=R.P. of e(2+2i)x1−4(1−D2+i2D4)x2
=R.P. of e(2+2i)x(1−D2+i2D4)−1−4x2
=R.P. of e(2+2i)x(1+D2+i2D4)+(D2+i2D4)2−4x2
=R.P. of e(2+2i)x(1+D2+i2D4)−D2−4x2
=R.P. of e(2+2i)x3D2−4−i4D16x2
=R.P. of e(2+2i)x6−4x2−i8x16
=e2x3−2x2cos2x+4xsin2x8
y=(C1x+C2)e2x+e2x3−2x2cos2x+4xsin2x8
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Elaine Dawe
Elaine Dawe, BMath, Mathematics & Computer Science, University of Waterloo (1985)
Answered Sep 11, 2018 · Author has 1.6k answers and 557.8k answer views
Just add '=0', and voilà, you’ve got a differential equation.
(D2−4D+4)y−x2e2xcos(2x)=0
Solve homogeneous equation:
(D2−4D+4)y=0
(D−2)2y=0
D=2
yh=C1e2x+C2xe2x
Solve non-homogeneous equation using Method of Variation of Parameters.
(D2−4D+4)y=x2e2xcos(2x)
y1=e2x
y2=xe2x
W(y1,y2)=y1y2′−y1′y2=(e2x)((2x+1)e2x)−(2e2x)(xe2x)=e4x
g(x)=x2e2xcos(2x)
u1=−∫y2g(x)W(y1,y2)dx=−∫xe2x⋅x2e2xcos(2x)e4xdx=−∫x3cos(2x)dx=(34x−12x3)sin(2x)+(38−34x2)cos(2x)
u2=∫y1g(x)W(y1,y2)dx=∫e2x⋅x2e2xcos(2x)e4xdx=∫x2cos(2x)dx=(12x2−14)sin(2x)+12xcos(2x)
yp=u1y1+u2y2=[(34x−12x3)sin(2x)+(38−34x2)cos(2x)]e2x+[(12x2−14)sin(2x)+12xcos(2x)]xe2x=12xe2xsin(2x)−(14x2−38)e2xcos(2x)
y=C1e2x+C2xe2x+12xe2xsin(2x)−(14x2−38)e2xcos(2x)
Answer:
Step-by-step explanation:
Particular Integral
[math]y_p=\dfrac 1 {(D-2)^2} x^2 e^{2x } \cos 2x[/math]
[math]=\text {R.P. of }\dfrac 1 {(D-2)^2} x^2 e^{(2+i)x }[/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac 1 {(D+2i)^2} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac 1 {D^2-4+i2D} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac 1 {-4(1-\frac {D^2+i2D}4)} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac {(1-\frac {D^2+i2D}4)^{-1}} {-4} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac {(1+\frac {D^2+i2D}4)+(\frac {D^2+i2D}4)^2} {-4} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac {(1+\frac {D^2+i2D}4)-D^2} {-4} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac {3D^2-4-i4D} {16} x^2 [/math]
[math]=\text {R.P. of }e^{(2+2i)x }\dfrac {6-4x^2-i8x} {16} [/math]
[math]=e^{2x }\dfrac {3-2x^2\cos 2x+4x\sin 2x} {8} [/math]
[math]y=\boxed{(C_1x+C_2)e^{2x}+e^{2x }\dfrac {3-2x^2\cos 2x+4x\sin 2x} {8} }[/math]