How Do You Solve Challenging Second-Order ODE Problems?

[tandoorichicken]

There are a couple of problems on this week's homework assignment that are giving me trouble.

(1) Find a particular solution for $y'' + y = t^2$ by using the method of undetermined coefficients

Here I don't know which coefficient expression to use, for example if the term on the right side was e^t, I could sub y = Ae^t so that all 'e^t's would cancel out and I would be left with an expression for A.

(2) Solve the IVP: $y'' - 4y' +2y = e^{2t}$, homogenous initial conditions at t=0.

What I did was what I normally do for any first-order ODE. I separated the problem out into homogenous and particular parts.
$$y_h: y''-4y'+2y=0$$
$$s^2-4s+2=0\rightarrow s=2\pm\sqrt{2}$$ where s is a characteristic root. Therefore $y_h=c_1 e^{(2+\sqrt{2})t}+c_2 e^{(2-\sqrt{2})t}$
For the particular part I used undetermined coefficients and subbed y=Ae^2t and got an expression for A: 4A - 8A + 2A = 1, so A = -1/2, and $y_p = -\frac{1}{2}e^{2t}$
What I am confused about is what comes next.
Is the solution then just $y = y_p + y_h = c_1 e^{(2+\sqrt{2})t}+c_2 e^{(2-\sqrt{2})t} -\frac{1}{2}e^{2t}$ ? And how do I account for the initial conditions?

 

[dextercioby]

Initial conditions help you fix the coefficients values.They are arbitrary,unless there are initial conditions.

For the first,substitute

$$y_{p}(t)=C(x) \sin x$$

Daniel.

 

[thepatientmental]

First of all, don't worry about feeling confused or struggling with certain problems in your homework. It's completely normal and part of the learning process. Just keep practicing and seeking help when needed.

To answer your first question about finding a particular solution for y'' + y = t^2, you can use the method of undetermined coefficients by considering the form of the right-hand side. Since t^2 is a polynomial of degree 2, you can assume a particular solution of the form y_p = At^2 + Bt + C. Then, substitute this into the equation and equate coefficients to find the values of A, B, and C.

For the second problem, you have correctly solved for the homogeneous solution and found the particular solution using undetermined coefficients. To account for the initial conditions, you can use the method of variation of parameters. This involves finding a particular solution in the form of y_p = u_1(t)y_1(t) + u_2(t)y_2(t), where y_1 and y_2 are the two linearly independent solutions of the homogeneous equation and u_1 and u_2 are functions to be determined. You can then use the initial conditions to solve for u_1 and u_2, and thus find the complete solution to the initial value problem.

I hope this helps clarify the next steps for solving these types of problems. Remember to always check your answers and seek help from your professor or classmates if needed. Keep up the good work!