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Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Find the solution to the following differential equations. y″ − 6y′ + 25y = 2 sin(t/2) – cos(t/2)   ?

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The auxiliary equation is r^2 - 6r + 25 = 0.  This yields roots of 3+4i and 3-4i.

The homogeneous solution is therefore y = e^3t(Acos(4t) + Bsin(4t)).

To this we must add the particular solution to get the general solution.

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

y = Asin(t/2) + Bcos(t/2)

y' = (1/2)Acos(t/2) - (1/2)Bsin(t/2)

y'' = -(1/4)Asin(t/2) - (1/4)Bcos(t/2)

Plugging these into the original equation gives:

(-1/4)Asin(t/2) - (1/4)Bcos(t/2) - 3Acos(t/2) + 3Bsin(t/2) +25Asin(t/2) + 25Bcos(t/2) = 2sin(t/2) - cos(t/2)

Combining like terms and equating terms on each side gives:

[-(1/4)A + 3B + 25A](sin(t/2)) + [-3A - (1/4)B + 25B](cos(t/2)) = 2sin(t/2) - cos(t/2)

[(99/4)A + 3B](sin(t/2)) + [-3A + (99/4)B](cos(t/2)) = 2sin(t/2) - cos(t/2)

A = 56/663 and B = -20/663  meaning the particular solution is:

y = (56/663)sin(t/2) - (20/663)cos(t/2).

The general solution is:

y = e^(3t)(Acos(4t) + Bsin(4t)) + (56/663)sin(t/2) - (20/663)cos(t/2), where A and B are constants to be determined by initial conditions.

• With explanations, the solution to this involves two sides of A4.

I have a solution written out, but fear that it will receive little thanks or acknowledgement.

Give me an encouraging reason to post.

:)>

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