To solve the given differential equation, we start by identifying the form of the equation and finding its characteristic equation.

The given differential equation is:

$y''' - 4y'' - 5y' = 0$

This is a linear homogeneous differential equation with constant coefficients. To solve it, we use the characteristic equation method.

Step 1: Form the Characteristic Equation

For a differential equation of the form:

$a_n y^{(n)} + a_{n-1} y^{(n-1)} + \ldots + a_1 y' + a_0 y = 0$

The characteristic equation is:

$a_n r^n + a_{n-1} r^{n-1} + \ldots + a_1 r + a_0 = 0$

In this case, the differential equation is:

$y''' - 4y'' - 5y' = 0$

This implies:

$1 \cdot y''' - 4 \cdot y'' - 5 \cdot y' + 0 \cdot y = 0$

The characteristic equation corresponding to this differential equation is:

$r^3 - 4r^2 - 5r = 0$

Step 2: Solve the Characteristic Equation

To solve the characteristic equation:

$r^3 - 4r^2 - 5r = 0$

we can factor out $r$:

$r(r^2 - 4r - 5) = 0$

This gives us:

$r = 0 \quad \text{or} \quad r^2 - 4r - 5 = 0$

Now, solve the quadratic equation:

$r^2 - 4r - 5 = 0$

Using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -4$, and $c = -5$:

$r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$

$r = \frac{4 \pm \sqrt{16 + 20}}{2}$

$r = \frac{4 \pm \sqrt{36}}{2}$

$r = \frac{4 \pm 6}{2}$

This gives us two solutions:

$r = \frac{4 + 6}{2} = 5 \quad \text{and} \quad r = \frac{4 - 6}{2} = -1$

Step 3: Write the General Solution

Now we have the roots $r = 0$, $r = 5$, and $r = -1$. Since all roots are distinct, the general solution to the differential equation is:

$y(t) = C_1 e^{0 \cdot t} + C_2 e^{5t} + C_3 e^{-t}$

$y(t) = C_1 + C_2 e^{5t} + C_3 e^{-t}$

where $C_1$, $C_2$, and $C_3$ are constants determined by initial conditions.

Summary

The general solution of the differential equation $y''' - 4y'' - 5y' = 0$ is:

$y(t) = C_1 + C_2 e^{5t} + C_3 e^{-t}$

Would you like more details on any step, or do you have any questions?

Related Questions:

1. How would you solve this differential equation with initial conditions?
2. What changes if the roots of the characteristic equation were complex?
3. How can the method of undetermined coefficients be used for non-homogeneous differential equations?
4. Can you solve a second-order linear differential equation using the characteristic equation method?
5. What if the differential equation had repeated roots?

Tip

For higher-order differential equations with constant coefficients, always start by writing the characteristic equation. Solving this equation is the key to finding the general solution.