Question

SOLVE. y''+3y'+2y=4e^x cos3x

Answer 1

Solve the homogeneous equation

`y'' + 3y' + 2y = 0`

Its characteristic equation is

`r^2 + 3r + 2 = (r + 1) (r + 2) = 0`

with roots at
`r=-1` and
`r=-2`, hence the characteristic solution is

`y_c = C_1 e^(-x) + C_2 e^(-2x)`

For the nonhomogeneous equation, I'll use variation of parameters. We're looking for a solution of the form

`y = u_1 y_1 + u_2 y_2`

to the equation

`y'' + a(x) y'' + b(x) y = f(x)`

such that

`\displaystyle u_1 = - \int (y_2f(x))/(W(y_1,y_2)) \, dx`

`\displaystyle u_2 = \int (y_1 f(x))/(W(y_1,y_2)) \, dx`

The Wronskian
`W(y_1,y_2)` of the two fundamental solutions
`y_1=e^(-x)` and
`y_2=e^(-2x)` is

`W(y_1,y_2) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}' \end{vmatrix} = -e^(-3x)`

Then we have

`\displaystyle u_1 = - \int (e^(-2x) \cdot 4e^x \cos(3x))/(-e^(-3x)) \, dx = 4 \int e^(2x) \cos(3x) \, dx`

`\displaystyle u_2 = \int (e^(-x) \cdot 4e^x \cos(3x))/(-e^(-3x)) \, dx = -4 \int e^(3x) \cos(3x) \, dx`

Recall Euler's identity,

`e^((a+bi)t) = e^(at) (\cos(bt) + i \sin(bt))`

Then we have the general antiderivative

`\displaystyle \int e^((a+bi)t) \, dt = \frac1{a+bi} e^((a+bi)t) + C = (a-bi)/(a^2+b^2) e^((a+bi)t) + C`

Taking the real parts of both sides, we have

`\displaystyle \mathrm{Re}\left\{\int e^((a+bi)t) \, dt \right\} = \mathrm{Re}\left\{(a-bi)/(a^2+b^2) e^((a+bi)t) + C\right\} \\\\ \int\,\mathrm{Re}\left\{e^((a+bi)t)\right\} \, dt = (e^(at))/(a^2+b^2) \mathrm{Re}\left\{(a-bi)(\cos(bt) + i \sin(bt))\right\} + C \\\\ \int e^(at) \cos(bt) \, dt = (e^(at))/(a^2+b^2) (a\cos(bt)+b\sin(bt)) + C`

so that

`\displaystyle u_1 = 4 \int e^(2x) \cos(3x) \, dx = (4e^(2x))/(13) (2\cos(3x) + 3 \sin(3x))`

and

`\displaystyle u_2 = -4 \int e^(3x) \cos(3x) \, dx = -\frac{2e^(3x)}3 (\cos(3x) + \sin(3x))`

We've found

`y = u_1 y_1 + u_2 y_2`

`\displaystyle y = (4e^x)/(13) (2\cos(3x) + 3 \sin(3x)) - \frac{2e^x}3 (\cos(3x) + \sin(3x))`

`\displaystyle y = \frac2{39} e^x (5\sin(3x) - \cos(3x))`

Then the general solution to the differential equation is

`\boxed{y(x) = C_1 e^(-x) + C_2 e^(-2x) + \frac2{39} e^x (5\sin(3x) - \cos(3x))}`