example solve the following differential


Example:  Solve the following differential equation.

y(3)  -  12 y''+48 y' + 64 y = 12 - 32 e-8t + 2 e4t

Solution:   

We first require the complementary solution thus the characteristic equation is,

r3 - 12 r2 + 48 r - 64 = (r - 4)3 = 0

r = 4; when multiply with 3

We've found a single root of multiplicity 3 thus the complementary solution is,

yc(t) = c1 e4t + c2 t e4t + c3 t2 e4t

Here is, our first guess for an exact solution is,

Yp = A + B e-8t + C e4t

Remember that the last term in our guess is into the complementary solution thus we'll require to add one at least one t to the third term in our guess. Also see that multiplying the third term with either t or t2 will result into a new term which is until now in the complementary solution and therefore we'll require multiplying the third term by t3 in order to find a term that is not included in the complementary solution.

Our last guess is,

Yp = A + B e-8t + C t3 e4t

Here all we require to do is take three derivatives of this, plug that in the differential equation and simplify to find (we'll leave this to you to verify the work now),

-64 A - 1728 B e-8t + 6 C e4t = 12 - 32 e -8t + 2 e4t

By setting coefficients equal and solving provides,

t0:         -64 A = 12        ⇒         A = -(3/16)

e-8t:      -1728 B = -32 ⇒         B = 1/54

e4t:       6C = 2              ⇒         C = 1/3

An exact solution is then,

Yp = -(3/16) + (1/54) e-8t + (1/3) t3 e4t

Then the general solution to above differential equation is,

y(t) = c1 e4t + c2 t e4t + c3 t2 e4t -(3/16) + (1/54) e-8t + (1/3) t3 e4t

Okay, we've only worked one example here, but remember that we mentioned earlier that with the exception of the extension to the method that we used in this example the work here is identical to work we did the 2nd order material.

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Application Programming: example solve the following differential
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