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Second order homogeneous

Second order homogeneous differential equations are crucial for modelling natural phenomena, like mechanical vibrations, electrical circuits and wave propagation. By learning how to solve these equations, you can understand and predict how systems behave under various conditions.

Second order homogeneous differential equations

A second order homogeneous differential equation is an equation that involves a function and its derivatives up to the second order. They are called "homogeneous" because they do not have any standalone constants; this means they are equal to \(0\) on the opposite side of the function and its derivatives.

The general form for a second order homogeneous equation is:

\[f(x)=a_{1}\dfrac{d^{2}y}{dx^{2}}+a_{2}\dfrac{dy}{dx}+a_{3}y\]

where \(a_{1}\), \(a_{2}\) and \(a_{3}\) are constants, and \(f(x)=0\).

Solving second order homogeneous differential equations

To solve second order homogeneous differential equations, we need an auxiliary equation. To generate an auxiliary equation, we let:

\[y''=\dfrac{d^{2}y}{dx^{2}}=m^{2}\] \[y'=\dfrac{dy}{dx}=m\] \[y=1\]

We substitute these values into our differential equation, then solve for \(m\).

Depending on our solutions for \(m\), we get a different complementary function, \(y_{c}x\), as shown in the table.

Solution to auxillary equation Complementary function
Two real and different solutions

\[m_{1}\textrm{ and }m_{2}\]

 \(y_{c}(x)=Ae^{m_{1}x}+Be^{m_{2}x}\)
One real, repeated solution

\[m_{1}=m_{2}\]

 \(y_{c}(x)=(At+B)e^{mx}\)
Complex solution

\[m=\alpha\pm\beta i\]

 \(y_{c}(x)=e^{\alpha x}(A\cos(\beta x)+B\sin(\beta x)\)

\(A\) and \(B\) are constants.

You can see that some solutions to the auxilliary equation may involve complex numbers, so make sure you are confident working with them!

Just like first order differential equations, we can solve for a general solution with a constant. To solve for a particular solution, we need to know initial or boundary values:

  1. \(y(0)\)
  2. \(\dfrac{dy}{dx}(0)\)

You can see how to find a particular solution in Example 2.

Example 1 – solving second order homogeneous differential equations

Solve for \(y(x)\) given \(2\dfrac{d^{2}y}{dx^{2}}-5\dfrac{dy}{dx}-3y=0\).

The auxiliary equation is:
\[2m^{2}-5m-3=0\]

We can solve for \(m\) by factorising.
\[\begin{align*} 2m^{2}-5m-3 & = 0\\
(2m+1)(m-3) & = 0\\
2m+1=0 & \textrm{ or } m-3=0\\
m_{1}=-\dfrac{1}{2} & \textrm{ or } m_{2}=3
\end{align*}\]

Next, we generate the complementary function. Here, we have two real and different solutions, so the complementary function is:
\[\begin{align*} y_{c}(x) & = Ae^{m_{1}x}+Be^{m_{2}x}\\
& = Ae^{-\frac{1}{2}x}+Be^{3x}
\end{align*}\]

where \(A\) and \(B\) are constants.

Find the solution to the differential equation \(6\dfrac{d^{2}y}{dx^{2}}+5\dfrac{dy}{dx}-4y=0\) with boundary conditions \(y(0)=11\) and \(\dfrac{dy}{dx}(0)=0\).

The auxiliary equation is:
\[6m^{2}+5m-4=0\]

We can factorise and solve for \(m\).
\[\begin{align*} 6m^{2}+5m-4 & = 0\\
(2m-1)(3m+4) & = 0\\
2m-1=0 & \textrm{ or } 3m+4=0\\
m_{1}=\frac{1}{2} & \textrm{ or }m_{2}=-\frac{4}{3}
\end{align*}\]

There are two real and different solutions for \(m\), so the complementary function is:
\[y_{c}(x)=Ae^{\frac{1}{2}x}+Be^{-\frac{4}{3}x}\]

To find constants \(A\) and \(B\), we need two simultaneous equations. For the first equation, we substitute \(y(0)=11\).
\[\begin{align*} 11 & = Ae^{0}+Be^{0}\\
A & = 11-B
\end{align*}\]

The other boundary condition we have is \(\dfrac{dy}{dx}(0)=0\), so to get the second equation, we need to differentiate the complementary function.
\[\frac{dy}{dx}=\frac{1}{2}Ae^{\frac{1}{2}x}-\frac{4}{3}Be^{-\frac{4}{3}x}\]

We substitute \(\dfrac{dy}{dx}(0)=0\) and get the second equation.
\[\begin{align*} 0 & = \frac{1}{2}Ae^{0}-\frac{4}{3}Be^{0}\\
B & = \frac{3}{8}A
\end{align*}\]

Remember, there are many ways to solve simultaneous equations. For this example, we will substitute the first equation into the second equation.
\[\begin{align*} A & = 11-\frac{3}{8}A\\
& = 8
\end{align*}\]

We substitute \(A=8\) back into the second equation to find \(B\).
\[\begin{align*} B & = \frac{3}{8}\times8\\
& = 3
\end{align*}\]

The final solution is:
\[y_{c}(x) = 8e^{\frac{1}{2}x}+3e^{-\frac{4}{3}x}\]

Exercise – solving second order homogeneous differential equations

  1. Find the general solutions for the following.
    1. \(\dfrac{d^{2}y}{dx^{2}}-6\dfrac{dy}{dx}+13y=0\)
    2. \(\dfrac{d^{2}y}{dx^{2}}-6\dfrac{dy}{dx}+9y=0\)
    3. \(2\dfrac{d^{2}y}{dx^{2}}-\dfrac{dy}{dx}-6y=0\)
    4. \(y''-10y'+25y=0\)
    5. \(y''+y'+y=0\)
  2. Find the particular solutions for the following.
    1. \(6\dfrac{d^{2}y}{dx^{2}}+5\dfrac{dy}{dx}-6y=0\) given that \(y=5\) and \(\dfrac{dy}{dx}=-1\) when \(x=0\)
    2. \(4\dfrac{d^{2}y}{dx^{2}}-5\dfrac{dy}{dx}+y=0\) given that \(y=1\) and \(\dfrac{dy}{dx}=-2\) when \(x=0\)
    3. \(x''-6x'+9x=0\) given that \(x(0)=2\) and \(x'(0)=0\)
    4. \(y''+6y'+13y=0\) given that \(y(0)=4\) and \(y'(0)=0\)

    1. \(y=e^{3x}(c\cos(2x)+D\sin(2x))\)
    2. \(y=Ae^{3x}+Bxe^{3x}\)
    3. \(y=Ae^{-\frac{3}{2}x}+Be^{2x}\)
    4. \(y=Ae^{5x}+Bxe^{5x}\)
    5. \(y=e^{-\frac{x}{2}} \left( A\cos \left(\dfrac{\sqrt{3}}{2}x \right) + A\sin \left( \dfrac{\sqrt{3}}{2}x \right) \right)\)
    1. \(y=3e^{\frac{2}{3}x}+2e^{-\frac{3}{2}x}\)
    2. \(y=4e^{\frac{1}{4}x}-3e^{x}\)
    3. \(x=2e^{3t}(1-3t)\)
    4. \(y=2e^{-3x}(2\cos(2x)+3\sin(2x))\)

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