dx2d2y+2dxdy+5y=xe−xcos2x
Solution of equation can be found in the form
y=y1+y2
Solution y1 can be found from equation
y′′+2y′+5y=0
k2+2k+5=0D=4−20=−16=16i2k1=2−2−4i=−1−2ik2=2−2+4i=−1+2iy1=e−x(C1cos2x+C2sin2x)
Solution y2 can be found in form
y2=xe−x((ax+b)cos2x+(cx+m)sin2x)==e−x((ax2+bx)cos2x+(cx2+mx)sin2x))y2′=e−xcos2x(x2(−a+2c)++x(2a−b+2m)+b)++e−xsin2x(x2(−2a−c)++x(−2b+2c−m)+m)y2′′=e−xcos2x(x2(−3a−4c)++x(−4a−3b−4m+8c)−2b+2a+4m)++e−xsin2x(x2(4a−3c)++x(−8a+4b−3m−4c)−4b−2m+2c)
Plug y2,y2′,y2′′ into the equation, then
e−xcos2x(x2(−3a−4c)++x(−4a−3b−4m+8c)−2b+2a+4m)++e−xsin2x(x2(4a−3c)++x(−8a+4b−3m−4c)−4b−2m+2c)++2e−xcos2x(x2(−a+2c)++x(2a−b+2m)+b)++2e−xsin2x(x2(−2a−c)++x(−2b+2c−m)+m)++5e−x((ax2+bx)cos2x+(cx2+mx)sin2x))==xe−xcos2xcos2x:x2:−3a−4c−2a+4c+5a=0⟹0=0x:−4a−3b−4m+8c+4a−2b+4m+5b=1⟹8c=1,c=81x0:−2b+2a+4m+2b=0⟹2a+4m=0sin2x:x2:4a−3c−4a−2c+5c=0⟹0=0x:−8a+4b−3m−4c−4b++4c−2m+5m=0⟹a=0x0:−4b−2m+2c+2m=0⟹−4b+2c=0b=161m=0
Then
y2=e−x(161xcos2x+81x2sin2x)
The solution of the equation is
y=e−x(C1cos2x+C2sin2x)++e−x(161xcos2x+81x2sin2x)
Comments
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Thank you very much for the great help.God bless you.
Dear Prashant Naik, thank you for correcting us.
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
The r.h.s is xe^-xcos2x.In your solution x is missing.