No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?dmoney123 said:Homework Statement
y''-y=t-4e^(-t)Homework Equations
method of undetermined coefficients
The Attempt at a Solution
solving for characteristic equation first
y''-y=0
r^2-1=0
c_1e^(-t)+c_2e^(t)
RHS
particular solution
t-4e^(-t)
y_p(t)= At+B+Ce^(-t)
dmoney123 said:y_pt'(t)=A-Ce^(-t)
y_p''(t)=Ce^(-t)
plug into LHS
Ce^(-t)-At-B-Ce^(-t)=t-4e^(-t)
-A=t
A=-1
B=0
C cancels on the left... I am not sure how to any further?
Thanks
Mark44 said:No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?
I'm not sure what you mean.dmoney123 said:You add a coefficient t
so the guess becomes At+B+Cte^(-t) right? When I plug that back in, it still cancels out
A 2nd order differential equation is a mathematical equation that involves a function, its first and second derivatives. It is typically used to model physical phenomena in fields such as physics, engineering, and economics.
Undetermined coefficients are the coefficients of the general solution of a 2nd order differential equation that are left unknown and need to be determined by solving the equation using initial or boundary conditions.
To solve a 2nd order differential equation using undetermined coefficients, first, find the general solution by equating the equation to zero and solving for the unknown coefficients. Then, use the initial or boundary conditions to determine the values of the coefficients and obtain the particular solution.
No, undetermined coefficients can only be used to solve 2nd order differential equations that have constant coefficients and non-homogeneous terms.
The limitations of using undetermined coefficients include the inability to solve 2nd order differential equations with variable coefficients or non-constant non-homogeneous terms. It also may not work for equations with repeated roots or complex roots.