In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te−2t−(8t+12) with initial values y(0)=−2 and y′(0)=1.

Question

Write the form of the particular solution and its derivatives. (Use A, B, C, etc. for undetermined coefficients.

Find

Y =

Y' =

Y" =

Answer 1

So the process for Undetermined Coefficients is to solve the characteristic equation for the homogeneous diff. eqn. first:

If you sub in y = e^rt y' = ry, and y'' = r²y

plugging in you get the quadratic that must be 0 if y is not trivially 0:

r² + 4r + 4 = 0 therefore r = -2

The homogeneous solution is Ae^-2t + Bte^-2t

The particular solution will from the undetermined coefficients will have the form

y_p = C₁t + C₂ to cover the polynomial terms

+ (C₃t +C₄)t²e^-2t (We had to bump up the t polynomial because we have a te^-2t in the homogeneous solution. In this situation you have to multiply the trial solution by tⁿ so that no term in the particular solution looks like the homogeneous solution.

Now you take the 1 st and 2nd derivative and put these through the equation. They will be equal to the right-hand side of 12te^-2t -8t -12 (I am assuming this is what you meant to write)

I guess you only have to find the coefficients C1 to C4 (just call them A to D) I have written Y for you, You can take the derivatives to get Y' and Y''

You can solve for the constants because each term is independent. (I don't know why the initial conditions are mentioned if you are not solving the whole problem)

Finally, you add the particular and homogeneous solutions and use the initial conditions to solve for the two constants of the homogeneous solution if you are solving the whole thing (which may not be the case here)

Solution from Wolfram Alpha without the initial conditions is y = c₁e^-2t + c₂te^-2t -2t³e^-2t - 2t -1 The particular solution arrived at is the last three terms.

Good luck!

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