Question

Find the general solutions of the following differential equations:

a) "y'' + 4y' + 4y = t^-2 e^-2t, t > 0"
b) "y'' + 4y = 3/sin(2t), 0 < t < π"

Answer 1

a) The general solution to the differential equation
`\(y'' + 4y' + 4y = t^(-2)e^(-2t), t > 0\) is \(y(t) = (A + Bt)e^(-2t) + (1)/(12)t^2e^(-2t)\)`, where
`\(A\)` and
`\(B\)` are arbitrary constants.

b) For the differential equation
`\(y'' + 4y = (3)/(\sin(2t)), 0 < t < \pi\)`, the general solution is
`\(y(t) = A\cos(2t) + B\sin(2t) - (3)/(4)\cot(2)t\)`, where
`\(A\)` and
`\(B\)` are arbitrary constants.

Step-by-step explanation:

a) To find the general solution for the first differential equation, we begin by solving the associated homogeneous equation
`\(y'' + 4y' + 4y = 0\),`which has solutions in the form
`\(y_h(t) = (A + Bt)e^(-2t)\).`

To find a particular solution for the non-homogeneous term
`\(t^(-2)e^(-2t)\),` we use the method of undetermined coefficients, assuming a particular solution of the form
`\(y_p(t) = Ct^2e^(-2t)\).` After finding
`\(y_h(t)\) and \(y_p(t)\)`, the general solution is the sum of these solutions, yielding
`\(y(t) = (A + Bt)e^(-2t) + (1)/(12)t^2e^(-2t)\)`, where
`\(A\)` and
`\(B\)` are arbitrary constants.

b) For the second differential equation, the associated homogeneous equation \
`(y'' + 4y = 0\)` has solutions
`\(y_h(t) = A\cos(2t) + B\sin(2t)\)`. To find a particular solution for the non-homogeneous term
`\((3)/(\sin(2t))\)`, we use the method of undetermined coefficients, assuming a particular solution of the form
`\(y_p(t) = -(3)/(4)\cot(2t)\)`. The general solution is the sum of
`\(y_h(t)\)` and
`\(y_p(t)\), resulting in \(y(t) = A\cos(2t) + B\sin(2t) - (3)/(4)\cot(2)t\)`, where
`\(A\)` and
`\(B\)`are arbitrary constants.