Solving non homogeneous differential equations
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section 4.5 #6,10,12,14,20,28,36,38
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(6) Find a general solution to the following non homogeneous differential equation:
y'' + 5y' + 6y = 6x^2 + 10x + 2 + 12 e^x; yp(x) = e^x + x^2
Decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equations
(a) y'' - y' + y = (e^t + t)^2
(b) y'' + y' + y = e^t + 7
(c) y'' - 2y' + 3y = cosh t
Find general solution to the following differential equations:
(20) y'' + 4y = sin q - cos q
(28) y'' + y' - 12 y = e^t + e2t - 1
(36) y'' - 4y' + 4y = t^2 e^t - e^(2t)
Find a particular solution to the given higher order differential equation:
(38) y^(4) - 5y'' + 4y = 10 cos t - 20 sin t
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Solution Summary
I have solved eight problems on differential equations. I have found general solutions and particular solutions to non homogeneous second order and higher order differential equations. I also have shown how to determine if the method of undetermined coefficients together with superposition can be utilized to solve given differential equations. This problem set contains a wide variety of problems encountered in upper level calculus course. This problem-solution set will be very useful in learning the solution process to differential equations and to prepare for upcoming examinations. Please download them for success in your calculus course.
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(6)
Corresponding Homogeneous equation:
y'' + 5 y' + 6y = 0
D^2 + 5D + 6 = 0
D = [-5 +/- sqrt(25 - 24)]/2 = -3 or -2
Hence the solution is a linear combination of e-3x and e-2x
yc = C1 e-3x + C2 e-2x
General Solution:
y(x) = ex + x2 + C1 e-3x + C2 e-2x
where C1 and C2 are arbitrary constants.
(10) (12) and (14) Answer is Yes to all three.
Explanations:
(10)-
Let the given function be Ly = (e^t + t)^2 where L is the operator d^/dx^2 - d/dx + 1
Right hand side can be expanded to be e^(2t) + 2 t e^t + t^2
One can find a particular solution to each of the individual differential equation,
Ly = e^(2t)
Ly = 2 t e^t
Ly = t^2
Let the particular solutions of the above DE be, y1, y2 and y3.
Hence the ...
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