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y" – y' – 2y = 2e-⁻ᵗ Use the method of variation of parameters to find a particular solution of the given differential equation.

User Bgrober
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Final answer:

To find a particular solution of the given differential equation using the method of variation of parameters, first find the general solution of the homogeneous equation. Then assume a particular solution of the form y_p = u_1(t)y_1 + u_2(t)y_2 and solve for the functions u_1(t) and u_2(t). Substitute the values of u_1(t) and u_2(t) back into the particular solution to obtain the solution of the given differential equation.

Step-by-step explanation:

To find a particular solution of the given differential equation using the method of variation of parameters, we first need to find the general solution of the homogeneous equation y'' - y' - 2y = 0. The characteristic equation is r^2 - r - 2 = 0, which factors as (r - 2)(r + 1) = 0. So the solutions are y_1 = e^2t and y_2 = e^(-t).

Next, we assume a particular solution of the form y_p = u_1(t)y_1 + u_2(t)y_2, where u_1(t) and u_2(t) are functions to be determined. We differentiate y_p to find y_p' and y_p'', and substitute them into the original differential equation.

By equating the coefficients of e^2t and e^(-t) to 0 and solving the resulting system of equations, we can find u_1(t) and u_2(t). Substituting the values of u_1(t) and u_2(t) back into the particular solution, we obtain the particular solution of the given differential equation.

User JvO
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