# Solving second-order homogeneous differential equations

## Homogeneous differential equations are equal to 0

Homogenous second-order differential equations are in the form

???ay''+by'+cy=0???

The differential equation is a second-order equation because it includes the second derivative of ???y???. It’s homogeneous because the right side is ???0???. If the right side of the equation is non-zero, the differential equation is called nonhomogeneous.

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

The first thing we want to learn about second-order homogeneous differential equations is how to find their general solutions. The formula we’ll use for the general solution will depend on the kinds of roots we find for the differential equation.

To find the roots, we’ll first make substitutions for the function ???y??? in terms of the variable ???r???. I like to say that the number of “hash marks” on ???y??? is equal to the exponent you’ll put on ???r???. In other words

Making these substitutions into the differential equation gives

???ar^2+br+c=0???

Once we’ve substituted, we have the standard form of a quadratic equation and we can factor the left side to solve for the roots of the equation.

The constants ???c_1??? and ???c_2??? remain in the general solution. Later on we’ll learn how to solve initial value problems for second-order homogeneous differential equations, in which we’ll be provided with initial conditions that will allow us to solve for the constants and find the particular solution for the differential equation.

## Examples of finding the general solution to a second-order homogeneous differential equation that has distinct real roots

## Take the course

### Want to learn more about Differential Equations? I have a step-by-step course for that. :)

## Finding the general solution for a differential equation with distinct real roots

**Example**

Find the general solution.

???y''+5y'+6y=0???

If we make substitutions for ???y??? in terms of ???r???, we get

???r^2+5r+6=0???

We’ll factor the left side and solve for ???r???.

???(r+2)(r+3)=0???

???r_1+2=0???

???r_1=-2???

and

???r_2+3=0???

???r_2=-3???

The roots are two real numbers that are unequal (they’re not equal to each other), so these are distinct real roots. Which means we’ll use the formula for the general solution for distinct real roots and get

???y(x)=c_1e^{-2x}+c_2e^{-3x}???

This is the general solution to the differential equation.

The differential equation is a second-order equation because it includes the second derivative of y. It’s homogeneous because the right side is 0.

## The general solution for a differential equation with equal real roots

**Example**

Find the general solution.

???y''+6y'+9y=0???

If we make substitutions for ???y??? in terms of ???r???, we get

???r^2+6r+9=0???

We’ll factor the left side and solve for ???r???.

???(r+3)(r+3)=0???

???r_1+3=0???

???r_1=-3???

and

???r_2+3=0???

???r_2=-3???

The roots are two real numbers that are equal (they’re equal to each other), so these are equal real roots. Which means we’ll use the formula for the general solution for equal real roots and get

???y(x)=c_1e^{-3x}+c_2xe^{-3x}???

This is the general solution to the differential equation.

We’ll do one last example where the roots of the differential equation are complex conjugate roots.

## Finding the general solution with complex conjugate roots

**Example**

Find the general solution.

???y''+2y'+17y=0???

If we make substitutions for ???y??? in terms of ???r???, we get

???r^2+2r+17=0???

Since we can’t factor the left side, we’ll use the quadratic formula to find the roots. We could also complete the square to get the same result.

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

???r=\frac{-(2)\pm\sqrt{(2)^2-4(1)(17)}}{2(1)}???

???r=\frac{-2\pm\sqrt{4-68}}{2}???

???r=\frac{-2\pm\sqrt{-64}}{2}???

???r=\frac{-2\pm\sqrt{(64)(-1)}}{2}???

???r=\frac{-2\pm8\sqrt{-1}}{2}???

Since the imaginary number ???i??? is defined as ???i=\sqrt{-1}???, we get

???r=\frac{-2\pm8i}{2}???

???r=-1\pm4i???

The roots are two complex numbers that are conjugates of one another, so these are complex conjugate roots. Which means we’ll use the formula for the general solution for complex conjugate roots. Matching these roots to ???r=\alpha\pm{\beta}i??? tells us that ???\alpha=-1??? and ???\beta=4???, so we get

???y(x)=e^{-1x}\left[c_1\cos{(4x)}+c_2\sin{(4x)}\right]???

???y(x)=e^{-x}\left[c_1\cos{(4x)}+c_2\sin{(4x)}\right]???

This is the general solution to the differential equation.