Question #105413
Using the method of undetermined coefficients, write the trial solution of the equation
d^2y/dx^2 + 2dy/dx +5y= xe^-x cos2x

and hence solve it.
1
Expert's answer
2020-08-19T17:01:34-0400

d2ydx2+2dydx+5y=xexcos2x\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=xe^{-x} \cos2x

Solution of equation can be found in the form

y=y1+y2y=y_1+y_2

Solution y1y_1 can be found from equation

y+2y+5y=0y''+2y'+5y=0

k2+2k+5=0D=420=16=16i2k1=24i2=12ik2=2+4i2=1+2iy1=ex(C1cos2x+C2sin2x)k^2+2k+5=0\\ D=4-20=-16=16i^2\\ k_1=\frac{-2-4i}{2}=-1-2i\\ k_2=\frac{-2+4i}{2}=-1+2i\\ y_1=e^{-x}(C_1\cos2x+C_2\sin2x)

Solution y2y_2 can be found in form

y2=xex((ax+b)cos2x+(cx+m)sin2x)==ex((ax2+bx)cos2x+(cx2+mx)sin2x))y2=excos2x(x2(a+2c)++x(2ab+2m)+b)++exsin2x(x2(2ac)++x(2b+2cm)+m)y2=excos2x(x2(3a4c)++x(4a3b4m+8c)2b+2a+4m)++exsin2x(x2(4a3c)++x(8a+4b3m4c)4b2m+2c)y_2=xe^{-x}((ax+b)\cos2x+(cx+m)\sin2x)=\\ =e^{-x}((ax^2+bx)\cos2x+(cx^2+mx)\sin2x))\\ y_2' =e^{-x}\cos2x(x^2(-a+2c)+\\+x(2a-b+2m)+b)+\\ +e^{-x}\sin2x(x^2(-2a-c)+\\+x(-2b+2c-m)+m)\\ y_2'' =e^{-x}\cos2x(x^2(-3a-4c)+\\+x(-4a-3b-4m+8c)-2b+2a+4m)+\\ +e^{-x}\sin2x(x^2(4a-3c)+\\+x(-8a+4b-3m-4c)-4b-2m+2c)\\

Plug y2,y2,y2y_2, y_2', y_2'' into the equation, then

excos2x(x2(3a4c)++x(4a3b4m+8c)2b+2a+4m)++exsin2x(x2(4a3c)++x(8a+4b3m4c)4b2m+2c)++2excos2x(x2(a+2c)++x(2ab+2m)+b)++2exsin2x(x2(2ac)++x(2b+2cm)+m)++5ex((ax2+bx)cos2x+(cx2+mx)sin2x))==xexcos2xcos2x:x2:3a4c2a+4c+5a=0    0=0x:4a3b4m+8c+4a2b+4m+5b=1    8c=1,c=18x0:2b+2a+4m+2b=0    2a+4m=0sin2x:x2:4a3c4a2c+5c=0    0=0x:8a+4b3m4c4b++4c2m+5m=0    a=0x0:4b2m+2c+2m=0    4b+2c=0b=116m=0e^{-x}\cos2x(x^2(-3a-4c)+\\+x(-4a-3b-4m+8c)-2b+2a+4m)+\\ +e^{-x}\sin2x(x^2(4a-3c)+\\+x(-8a+4b-3m-4c)-4b-2m+2c)+\\ +2e^{-x}\cos2x(x^2(-a+2c)+\\+x(2a-b+2m)+b)+\\ +2e^{-x}\sin2x(x^2(-2a-c)+\\+x(-2b+2c-m)+m)+\\ +5e^{-x}((ax^2+bx)\cos2x+(cx^2+mx)\sin2x))=\\ =xe^{-x}\cos2x\\ \cos2x:\\ x^2:-3a-4c-2a+4c+5a=0\\\implies0=0\\ x:-4a-3b-4m+8c+4a-2b+4m+5b=1\\ \implies8c=1, c=\frac{1}{8}\\ x^0:-2b+2a+4m+2b=0\\ \implies2a+4m=0\\ \sin2x:\\ x^2:4a-3c-4a-2c+5c=0\\ \implies0=0\\ x:-8a+4b-3m-4c-4b+\\+4c-2m+5m=0\\ \implies a=0\\ x^0:-4b-2m+2c+2m=0\\ \implies-4b+2c=0\\ b=\frac{1}{16}\\ m=0

Then

y2=ex(116xcos2x+18x2sin2x)y_2=e^{-x}(\frac{1}{16}x\cos2x+\frac{1}{8}x^2\sin2x)

The solution of the equation is

y=ex(C1cos2x+C2sin2x)++ex(116xcos2x+18x2sin2x)y=e^{-x}(C_1\cos2x+C_2\sin2x)+\\ +e^{-x}(\frac{1}{16}x\cos2x+\frac{1}{8}x^2\sin2x)



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Comments

Assignment Expert
25.08.20, 16:41

Dear Prashant Naik, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Prashant Naik
25.08.20, 16:36

Thank you very much for the great help.God bless you.

Assignment Expert
17.08.20, 21:39

Dear Prashant Naik, thank you for correcting us.

Prashant Naik
17.08.20, 16:23

I see x is missing on the RHS of the equation.As the process of getting values of constants is very tedious request for your kind help to solve the given correct equation

P Naik
17.08.20, 12:34

The r.h.s is xe^-xcos2x.In your solution x is missing.

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