2nd order inhomogenous equations

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SUMMARY

The forum discussion centers on solving the second-order inhomogeneous differential equation \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = g(x)\), where \(g(x) = -14 \cos(2x) - 2 \sin(2x)\). The homogeneous solution is identified as \(y = e^x (A \cos(x) + B \sin(x))\). For the particular integral, the correct approach is to use \(y_p = C \sin(2x) + D \cos(2x)\), rather than including additional terms. The general solution combines both the homogeneous and particular solutions.

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adichy
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Homework Statement



Find the general solution to the differential equation
[tex]\frac{d^2y}{dx^2}[/tex] -2*[tex]\frac{dy}{dx}[/tex] +2y =g(x)

where g(x) = −14 cos(2x) − 2 sin(2x)

Homework Equations





The Attempt at a Solution


ive found the homogenous solution which is
y=e^x (ae^ix +be^-ix)

what I am not sure of is for the particular intergral do i do
Csin(2x)+Dcos(2x)+Esin(2x)+Fcos(2x)
or is it just Csin(2x)+Dcos(2x)
please advise

thanks

edit:cant seem to get the latex to come out right basically its d^2y/dx^2 - 2(dy/dx) + 2y=0
 
Last edited:
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Tip: Use only one pair of tex tags, one at the beginning of the equation and one at the end.
adichy said:

Homework Statement



Find the general solution to the differential equation
[tex]\frac{d^2y}{dx^2}[/tex] -2*[tex]\frac{dy}{dx}[/tex] +2y =g(x)

where g(x) = −14 cos(2x) − 2 sin(2x)

Homework Equations





The Attempt at a Solution


ive found the homogenous solution which is
y=e^x (ae^ix +be^-ix)
You can also write this as y = ex(A cos(x) + B sin(x))
adichy said:
what I am not sure of is for the particular intergral do i do
Csin(2x)+Dcos(2x)+Esin(2x)+Fcos(2x)
or is it just Csin(2x)+Dcos(2x)
For your particular solution, use yp = Csin(2x)+Dcos(2x)
adichy said:
please advise

thanks

edit:cant seem to get the latex to come out right basically its d^2y/dx^2 - 2(dy/dx) + 2y=0
 
I would start with a particular solution of:
[tex] P.I.=A\cos 2x+B\sin 2x[/tex]
and the general solution is:
[tex] y=Ce^{\lambda_{1}x}+De^{\lambda_{2}x}[/tex]
where the lambda are solutions of:
[tex] \lambda^{2}-2\lambda+2=0[/tex]
 

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