Differential Equations Exercise 26

We want to solve the following differential equation:

$$ \begin{cases} y''+6y'+8y = 0 \\ y(0)=-2 \\ y'(0)=1 \end{cases} $$

This is a second-order homogeneous differential equation with two initial conditions (a Cauchy problem).

To begin, we form the characteristic equation using the auxiliary variable $z$:

$$ z^2+6z+8=0 $$

$$ z = \frac{-6 \pm \sqrt{36-32}}{2} = \frac{-6 \pm 2}{2} = \begin{cases} z=\frac{-6-2}{2}=-4 \\ \\z=\frac{-6+2}{2}=-2 \end{cases} $$

The characteristic polynomial has two distinct real roots: $z_1=-4$ and $z_2=-2$.

Therefore, the general solution of the differential equation is:

$$ y = c_1 e^{-4x} + c_2 e^{-2x} $$

Differentiating, we obtain:

$$ y' = -4c_1 e^{-4x} - 2c_2 e^{-2x} $$

This gives us the system:

$$ \begin{cases} y = c_1 e^{-4x} + c_2 e^{-2x} \\ \\ y' = -4 c_1 e^{-4x} - 2 c_2 e^{-2x} \end{cases} $$

Now we apply the initial conditions $x=0,\, y=-2,\, y'=1$ to the general solution:

$$ \begin{cases} -2 = c_1 e^0 + c_2 e^0 \\ \\ 1 = -4 c_1 e^0 - 2 c_2 e^0 \end{cases} $$

which simplifies to:

$$ \begin{cases} -2 = c_1 + c_2 \\ \\ 1 = -4c_1 - 2c_2 \end{cases} $$

We solve this system by substitution:

$$ \begin{cases} c_1 =-2-c_2 \\ \\ 1 = -4(-2-c_2) - 2c_2 \end{cases} $$

$$ \begin{cases} c_1 =-2-c_2 \\ \\ 1 = 8 + 4c_2 - 2c_2 \end{cases} $$

$$ \begin{cases} c_1 =-2-c_2 \\ \\ -7 = 2c_2 \end{cases} $$

$$ \begin{cases} c_1 =-2-c_2 \\ \\ c_2 = -\tfrac{7}{2} \end{cases} $$

$$ \begin{cases} c_1 = -2 - \left(-\tfrac{7}{2}\right) \\ \\ c_2 = -\tfrac{7}{2} \end{cases} $$

$$ \begin{cases} c_1 = \tfrac{-4+7}{2} \\ \\ c_2 = -\tfrac{7}{2} \end{cases} $$

$$ \begin{cases} c_1 = \tfrac{3}{2} \\ \\ c_2 = -\tfrac{7}{2} \end{cases} $$

Having determined the constants $c_1 = \tfrac{3}{2}$ and $c_2 = -\tfrac{7}{2}$, we substitute them back into the general solution:

$$ y = \tfrac{3}{2} e^{-4x} - \tfrac{7}{2} e^{-2x} $$

This is the unique solution to the given Cauchy problem.