Question

Solve the given differential equation by undetermined coefficients y"+3y'+2y=6

Answer 1

`\[y(t) = C_1 e^(-t) + C_2 e^(-2t) + 2t - 2\]`

Step-by-step explanation:

The given differential equation is \(y'' + 3y' + 2y = 6\). To solve this non-homogeneous linear differential equation, we assume a particular solution of the form \(y_p = At + B\), where \(A\) and \(B\) are constants. Taking the first and second derivatives, we find
`\(y'_p = A\) and \(y''_p = 0\).`Substituting these into the original equation, we get \(0 + 3A + 2(At + B) = 6\), which simplifies to
`\(3A + 2At + 2B = 6\)`. Equating coefficients, we find
`\(A = 2\) and \(B = -2\)`.

The homogeneous solution can be found by solving the associated homogeneous equatio
`n \(y''_h + 3y'_h + 2y_h = 0\)`. The characteristic equation is
`\(r^2 + 3r + 2 = 0\)`, which factors as
`\((r + 1)(r + 2) = 0\).` Thus, the homogeneous solution is
`\(y_h = C_1e^(-t) + C_2e^(-2t)\)` where
`\(C_1\) and \(C_2\`) are arbitrary constants.

Combining the homogeneous and particular solutions, we get the general solution
`\(y(t) = y_h + y_p = C_1e^(-t) + C_2e^(-2t) + 2t - 2\)`, where
`\(C_1\)`and
`\(C_2\)` are determined by initial conditions if provided.