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If I let w=y', would I be able to solve this differential equation? I'm currently stuck and I just want to know if making this substitution is why I am stuck.

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- Thread starter Turion
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- #1

- 143

- 2

If I let w=y', would I be able to solve this differential equation? I'm currently stuck and I just want to know if making this substitution is why I am stuck.

- #2

SteamKing

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First, solve the homogeneous equation.

Second, find the particular solution.

Third, add the particular solution to the complementary solution (of the homogeneous equation).

- #3

arildno

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Yes, that works nicely.

If I let w=y', would I be able to solve this differential equation? I'm currently stuck and I just want to know if making this substitution is why I am stuck.

If we look at the homogenous system, you get the characteristic polynomial, with e^rx as trial solution:

r^2-5r+6=0, giving r=3 and r=2 as possibles. Furthermore, you have w_p1=4/3, for the constant particular solution.

Now, in order to solve for the particular solution 2sinx, you generally will need w_p=Asin(x)+Bcos(x)

Inserting this gives you the two equations in A and B

sin(x): -A+5B+6A=2 goes to: 5A+5B=2

cos(x):-B-5A+6B=0 goes to: 5B-5A=0

Thus, A=B=1/5.

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