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January 31st, 2012, 03:51 PM   #1
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Ordinary differential equation y' = 2y + 4

Hey guys, I have to solve this by finding a constant solution and writing y(t) = y_h(t) + y_p(t).

So far I have:

y' = 2y + 4
y_p = at + b
(at + b)' - 2(at + b) = 4
a - 2at - 2b = 4.

Now I am stuck. Also, my notes say to set y' = 0, but I am not sure what that means exactly.

Any help?
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January 31st, 2012, 05:23 PM   #2
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Re: Ordinary differential equation y' = 2y + 4

We are given the ODE:



If I were after the general solution, I would first arrange it as:



then multiply through by the integrating factor :







Integrate with respect to t:







However, you are instructed to find the solution to the corresponding homogeneous equation and a particular solution:



The associated auxiliary equation is:



Thus:



Using the method of undetermined coefficients, we assume a particular solution of the form:

where A is an arbitrary real constant, and thus:



Substituting into the ODE gives:





Thus, by the principle of superposition, we have:

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January 31st, 2012, 11:57 PM   #3
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Re: Ordinary differential equation y' = 2y + 4

Quote:
Originally Posted by Norm850
Hey guys, I have to solve this by finding a constant solution and writing y(t) = y_h(t) + y_p(t).

So far I have:

y' = 2y + 4
y_p = at + b
(at + b)' - 2(at + b) = 4
a - 2at - 2b = 4.

Now I am stuck. Also, my notes say to set y' = 0, but I am not sure what that means exactly.

Any help?
a - 2at - 2b = 4. any value of t. So a = 0 and -2b=4
But this gives ony one solution of the ODE, not all.

dy/dx=2y+4 = 2(y+2)
dy/(y+2) ) =2 dx
ln(y+2)=2x+constant
y=C*exp(2x) -2
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