what is the y-particular of:
y"-2y'+2y=(e^t)(sin(t))
To find the particular solution (y-particular) of the given differential equation, we can use the method of undetermined coefficients.
Step 1: Find the complementary solution (y-complement) of the homogeneous equation.
The homogeneous equation is y"-2y'+2y=0. To find the complementary solution, we assume a trial solution of the form y = e^(mt). Plugging this into the homogeneous equation, we get the characteristic equation:
(m^2 - 2m + 2)e^(mt) = 0.
The characteristic equation (m^2 - 2m + 2) = 0 has complex roots m = 1 ± i.
Therefore, y-complement = C1 e^(t) cos(t) + C2 e^(t) sin(t), where C1 and C2 are constants.
Step 2: Determine the particular solution.
Since the right-hand side of the differential equation contains terms of the form e^t sin(t), our guess for the particular solution would be of the form y-particular = Ate^t cos(t) + Bte^t sin(t), where A and B are coefficients to be determined.
Step 3: Substitute the guess into the differential equation and solve for coefficients.
Plug y-particular into the differential equation y"-2y'+2y=(e^t)(sin(t)). Derive y-particular twice using the product rule and rearrange terms to simplify.
The resulting equation will have terms involving e^t cos(t) and e^t sin(t), as well as their derivatives with respect to t. Equate the coefficients of these terms to the corresponding terms on the right-hand side of the original equation.
This will give you a system of linear equations in terms of A and B. Solve this system to find the values of A and B.
Step 4: Write the particular solution.
Once you have determined the values of A and B, substitute them back into the particular solution y-particular = Ate^t cos(t) + Bte^t sin(t). This gives you the final expression for the particular solution.
Note: The particular solution obtained will be unique for the given differential equation.
I hope this explanation helps you understand the steps involved in finding the y-particular of the given differential equation.