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Section 3.4 Reduction of Order
Solve for the characteristic equation.
If the roots are r1,r2
y = C
1
e
r1t
+ C
2
e
r2t
R1 and r2 complex roots, \ +iu
y = C
1
e
\ t
cos(ut) + C
2
e
\ t
sin(ut)
R1=R2
y = C
1
e
r1t
+ C
2
te
r2t
Let y = v(t)*(solution already given)
Take derivatives y’ and y’’.
Plugging into the original equation.
Using that t > 0 we use the equation:
V’’ = 0 integrating twice we get:
V = c1t + c2
Use c1 = 1, c2 = 0 for best C’s possible
Plug v(t) into y(t)
Second solution is found to be y(t)
Section 3.5 Undetermined Coefficients
The characteristic polynomial has roots # and #
The solutions to the homogenous/complimentary problem are:
Y
h
= c
1
e
3t
+ C
2
e
t
To find a particular solution we try the form:
Y(t) = Acos(2t) + Bsin(2t)
And take the first and second derivatives
Plug into the original equation.
Solve for A and B, getting rid of “ghosts”.
Once we solve for A and B, we add the particular and the
homogenous/complimentary equations.
For Equations with a g(x) = 3te
t
Try Ate
t
, then Ate
t
+ Bt
2
e
t
Section 3.6 Variations of Parameters
First we consider the homogenous equation:
Y’’ + 4y’ + 4y = 0
Using the characteristic equation:
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2
+ 4R + 4 = 0
Solve for R.
Then its complimentary solution is
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 Spring '11
 DELALLAVE
 Differential Equations, Equations

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