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Homogeneous Differential Equation – Definition, Solutions, and Examples
Understanding how to work with homogeneous differential equations is important if we want to explore more complex calculus topics and work on advanced endeavors in other disciplines such as physics, mathematics, and finance.
Homogeneous differential equations are differential equations where each term will be of the form, $y^{(n)}q(x)$. Oftentimes, youâ€™ll see homogeneous equations having zero on the right-hand side of their equation.
This article gives you a comprehensive idea of what makes homogeneous differential equations unique. Weâ€™ll also summarize the important techniques youâ€™ll need to work on different types of homogeneous differential equations. For now, letâ€™s begin by making sure that we know how to identify and rewrite homogeneous differential equations.
What Is a Homogeneous Differential Equation?
A homogeneous differential equation contains a differential expression on its left-hand side and zero on the right-hand side of the equation. Using this definition, the general form of a homogeneous differential equation is as shown below.
\begin{aligned}a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +â€¦+a_1y^{\prime}+ a_0y &= 0\end{aligned}Â
In this general form, $a_n(x), a_{n -1}(x), â€¦, a_1(x), a_0(x)$, are factors in terms of $x$. When the homogeneous differential equation is also linear, these become constant factors, $a_n, a_{n-1}, â€¦,a_1, a_0$. In the past, weâ€™ve also learned about first order and second order differential equations. Below are their general forms as we have learned before:
First Order Homogeneous Differential Equation | \begin{aligned}\dfrac{dy}{dx} &= f(x, y)\\ P(x, y) \phantom{x}dx + Q(x, y) \phantom{x}dy &= 0\end{aligned} |
Second Order Homogeneous Differential Equation | \begin{aligned}y^{\prime \prime} + P(x)y^{\prime} + Q(x)y &= f(x)\\\dfrac{d^2y}{dx^2}+ P(x)\dfrac{dy}{dx} + Q(x)y &= f(x) \end{aligned} |
Use these definitions and the general form of the homogeneous differential equations to classify them. Knowing how to identify homogeneous differential equations is important since techniques for solving differential equations would depend on whether they are homogeneous or not.
How To Tell if a Differential Equation Is Homogeneous
Suppose that we have a differential equation, $a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +â€¦+a_1y^{\prime}+ a_0y =g(x)$, we can identify whether it is homogeneous or not depending on the value of $g(x)$.
- When we can show that $g(x) = 0$, the differential equation is homogeneous.
- Otherwise, the equation is non-homogeneous.
Let us show you two examples to demonstrate how a differential equation looks when it is homogeneous and when it is non-homogeneous.
\begin{aligned}\boldsymbol{y^{\prime \prime} \cos x â€“ y\sinÂ x = y^{\prime}}\end{aligned}
We can rewrite the equation so that all terms with $y$ and its derivatives are on the left-hand side.
\begin{aligned} y^{\prime \prime} \cos x â€“ y\sinÂ x &= y^{\prime}\\ y^{\prime \prime} \cos x -y^{\prime} â€“ y\sinÂ x &=0 \end{aligned}
Since weâ€™ve written the differential equation in its standard form, we have shown that $g(x) = 0$, so our equation is indeed homogeneous. Now, letâ€™s show you an example of a non-homogeneous equation for comparison.
\begin{aligned}\boldsymbol{y^{\prime \prime} \cos x +Â y\cosÂ x â€“ 5xe^x = 0}\end{aligned}
Â Once again, we rewrite the differential equation in the form of $a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +â€¦+a_1y^{\prime}+ a_0y =g(x)$ and see if $g(x)$ is zero or not.
\begin{aligned}y^{\prime \prime} \cos x +Â y\cosÂ x â€“ 5xe^x &= 0\\ y^{\prime \prime} \cos x +Â y\cosÂ x &= 5xe^x \end{aligned}
Since $g(x) = 5xe^x$ and itâ€™s not equal to zero, the equation is not homogeneous, or a non-homogeneous differential equation. Now that we know how to identify homogeneous equations, itâ€™s time for us to refresh our knowledge on the different techniques of finding the solutions of homogeneous differential equations.
How To Solve Homogeneous Differential Equations?
There are different ways for us to solve homogeneous differential equations and itâ€™s important that we know the common techniques used when dealing with them. First, letâ€™s lay out the methods we have for first order homogeneous differential equations.
\begin{aligned} P(x, y)\phantom{x}dx + Q(x, y) \phantom{x}dy = 0$
- When we can show that the first order differential equation is an exact equation, we use the appropriate technique of solving exact equations. The solution of the exact equation will be as shown below.
\begin{aligned} f(x, y) &= \int P(x, y) \phantom{x}dx +g(y)\\f(x, y) &= \int Q(x, y) \phantom{x}dy +g(x)\end{aligned}
- When the first linear homogeneous linear equation is not exact, we can use the integrating factor Multiply both sides of the equation by the integrating factor to solve the now exact equation.
\begin{aligned}\boldsymbol{\mu(x)}\end{aligned} | \begin{aligned}\boldsymbol{\mu(y)}\end{aligned} |
\begin{aligned}\mu&= \text{exp}\left[\int\dfrac{ \left(\partial P/\partial y -\partial Q/\partial x \right )}{Q} \phantom{x}dx\right ]\end{aligned} | \begin{aligned}\mu&= \text{exp} \left[\int\dfrac{ \left(\partial Q/\partial x – \partial P/\partial y\right )}{P} \phantom{x}dy \right ] \end{aligned} |
Now, when working with second order differential equations, we use the characteristic equation of the second order homogeneous differential equation. Once we know the nature of the characteristic equationâ€™s roots, use the chart below to help guide you in determining the solutionâ€™s general form.
Rootsâ€™ Nature | Discriminant | Solutionâ€™s General Form |
When the roots are real and distinct. | \begin{aligned}b^2 -4ac > 0 \end{aligned} | \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x} \end{aligned} |
When the two real roots are equal. \begin{aligned}r_1 = r_2 = r \end{aligned} | \begin{aligned}b^2 -4ac = 0 \end{aligned} | \begin{aligned}y(x) &= e^{rx} (C_1 + C_2 x) \end{aligned} |
When the resulting roots are complex. \begin{aligned}r_1 &= \alpha + \beta i\\ r_2 &= \alpha – \beta i\end{aligned} | \begin{aligned}b^2 -4ac < 0 \end{aligned} | \begin{aligned}y(x) &= e^{\alpha x} [C_1 \cos (\beta x) + C_2 \sin (\beta x)]\end{aligned} |
We can actually apply a similar process when working with higher order homogeneous differential equations. This time, however, weâ€™re working with more than two roots so weâ€™ll break down for you the possible general forms of the homogeneous differential equation with higher order.
Rootsâ€™ Nature | Solutionâ€™s General Form |
When the roots are real and distinct. | \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x} +â€¦+ C_ne^{r_n x} \end{aligned} |
When multiplicity exists within the roots. $m$ roots: $\{r_1, r_2, â€¦,r_m\}$ The multiplicity may vary, so letâ€™s generalize the multiplicity as: multiplicity: $\{k_1, k_2, â€¦,k_m\}$ Â Â | \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2xe^{r_1 x} + …+C_{k_1}x^{k_1 – 1}e^{r_1 x}+…\\&+C_{n -k_m +1}e^{r_m x} +C_{n -k_{m}+ 2}xe^{r_m x} + …+C_nx^{k_m -1}e^{r_m x}\end{aligned} |
When the resulting roots are complex and unique. \begin{aligned}r_{1,2} &= \alpha + \beta i\\ r_{3,4 }&= \gamma- \delta i\\&.\\&.\\&.\end{aligned} | \begin{aligned}y= {e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {e^{\gamma x}}\left( {{C_3}\cos \delta x + {C_4}\sin \delta x} \right) + \cdots\end{aligned} |
As you can see, the approach to finding the solutions to higher order homogeneous differential equations is quite similar to previous techniques. Donâ€™t worry, weâ€™ve provided enough examples for you to try working with differential equations with higher order.
If the homogeneous differential equation also includes initial conditions, we can solve the initial value problem by substituting the general solution into the equation. Use the resulting equation to find the expression for constants. When youâ€™re ready to work with different homogeneous differential equations, head over to the section below and start solving different types of homogeneous differential equations!
Example 1
Find the general solution of the homogeneous differential equation, $y^{\prime\prime} -3y^{\prime} + 2y = 0$. What would the particular solution be if we have the following initial conditions: $y(0) = 4$ and $y^{\prime}(0) = 2$.
Solution
Weâ€™re given a second order homogeneous differential equation, so letâ€™s use its characteristic equation and solve for the roots of the resulting quadratic equation.
\begin{aligned}y^{\prime\prime} -3y^{\prime} + 2y &= 0\\r^2 – 3r + 2&= 0\\(r -1)(r -2)&= 0\\r_1 =1&, r_2 =2\end{aligned}
The two roots are unique and real, so the general form of the differential equation is equal to $y(x) = C_1e^{r_1 x} + C_2e^{r_2 x}$.
\begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x}\\y(x) &= C_1e^{1 x} + C_2e^{2x}\end{aligned}
This means that the general solution of the equation is $y = C_1e^{x} + C_2e^{2x}$. Now, letâ€™s deal with the initial-value problem given that $y(0) = 4$ and $y^{\prime}(0) = 2$.
\begin{aligned}y &= C_1e^{x} + C_2e^{2x}\\y(0) &= C_1e^{0} + C_2e^{2 \cdot 0}\\4&= C_1+ C_2\end{aligned} | \begin{aligned}y^{\prime} &=\dfrac{d}{dx}\left( C_1e^{x} + C_2e^{2x} \right )\\y^{\prime} &= C_1e^x +2C_2 e^{2x}\\y^{\prime}(0) &= C_1e^{0} + 2C_2e^{2 \cdot 0}\\2&= C_1+ 2C_2\end{aligned} |
Use these two equations to find $C_1$ and $C_2$. You should end up with $C_1 = 6$ and $C_2 = -2$. Hence, if we have the given initial conditions, $y(0) = 4$ and $y^{\prime}(0) = 2$, our particular solution would be equal to $y = 6e^{x}Â – 2e^{2x}$.
Example 2
Solve the initial value problem, $y^{\prime\prime\prime} â€“ 2y^{\prime \prime} â€“ 5y^{\prime} + 6y = 0$, with following initial conditions:
\begin{aligned} y(0)&= 2\\ y^{\prime}(0) &= -4\\y^{\prime \prime}(0) &= -6\end{aligned}
Solution
The differential equation is already in the general form for homogeneous equations, so we can immediately work with the characteristic equation and solve for the roots of the resulting cubic equation.
\begin{aligned}y^{\prime\prime\prime} – 2y^{\prime \prime} – 5y^{\prime} + 6y &= 0\\r^3 – 2r^2 – 5r + 6&=0\\ (r + 2)(r – 1)(r – 3) &= 0\\\\r_1 = -2, r_2 = 1,r_3 &= 3 \end{aligned}
The three roots are all distinct and unique, so the homogeneous differential equationâ€™s general solution will have a form of $ y(x) = C_1e^{r_1 x} + C_2e^{r_2 x}+ C_3e^{r_3 x}$.
\begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x} + C_3e^{r_3 x}\\y(x) &= C_1e^{-2x} + C_2e^{x} + C_3e^{3x}\end{aligned}
Use the general solution and find the expressions for $y^{\prime}$ and $y^{\prime \prime}$.Â
\begin{aligned}y &= C_1e^{-2x} + C_2e^{x} + C_3e^{3x}\\y^{\prime} &= -2C_1e^{-2x} + C_2e^{x} + 3C_3e^{3x}\\y^{\prime\prime} &= 4C_1e^{-2x} + C_2e^{x} + 9C_3e^{3x}\end{aligned}
Substitute the initial conditions into the equations and set up a system of linear equations.
\begin{aligned}2 =y(0)&= C_1+ C_2 + C_3\\-4 = y^{\prime}(0)&= -2C_1 + C_2 + 3C_3\\-6 =y^{\prime \prime}(0)&= 4C_1 + C_2 + 9C_3\end{aligned}
Apply appropriate algebraic technique (or an equation solver) to solve this system of linear equations. This results to the following constants: $C_1 = \dfrac{16}{15}$, $C_2 = \dfrac{7}{3}$, and $C_3 = -\dfrac{7}{5}$.
\begin{aligned} y &= C_1e^{-2x} + C_2e^{x} + C_3e^{3x}\\&= \dfrac{16}{15}e^{-2x} +\dfrac{7}{3}e^x -\dfrac{7}{5}e^{3x} \end{aligned}
We now have the particular solution for our initial value problem: $y = \dfrac{16}{15}e^{-2x} +\dfrac{7}{3}e^x -\dfrac{7}{5}e^{3x}$.
Example 3
Find the general solution of the differential equation, $y^{(5)} â€“ 4y^{(4)} + 14y^{\prime\prime\prime} = 20y^{\prime\prime} -25y $.
Solution
Rewrite the differential equation first so that they exhibit the general form of the homogeneous differential equation.
\begin{aligned} y^{(5)} â€“ 4y^{(4)} + 14y^{\prime\prime\prime} &= 20y^{\prime\prime} -25y \\ y^{(5)} â€“ 4y^{(4)} + 14y^{\prime\prime\prime} – 20y^{\prime\prime} +25y&= 0 \end{aligned}
Now, use its characteristic equation to solve for its roots as shown below.
\begin{aligned} r^5 – 4r^4 + 14r^3 – 20r^2 +25r&= 0\\r(r^4 – 4r^3 + 14r^2 – 20r +25)&= 0\\r(r^2 -2r +5)^2&= 0\\r[r -(1 + 2i)]^2[r -(1 -2i)]^2&= 0\\\\r =0, r = 1+ 2i, r =1 &-2i\end{aligned}
For the first root, $r_1 = 0$, the solution will have a constant of $C_1$. The rest of the roots are unique and complex, so we can use the general form, $y= {e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {e^{\gamma x}}\left( {{C_3}\cos \delta x + {C_4}\sin \delta x} \right) $ to guide us in writing the rest of the components. Keep in mind though that each complex root has a multiplicity of two, so the last two terms will help a multiple of $x$.
\begin{aligned}y&= C_1 + \left(C_2e^{-x} \cos 2x + C_3e^{-x}\sin 2x \right ) +\left(C_4xe^{-x} \cos 2x + C_5xe^{-x}\sin 2x \right ) \end{aligned}
Practice Questions
1. Find the general solution of the homogeneous differential equation, $y^{\prime\prime} – 5y^{\prime} + 6y = 0$. What would the particular solution be if we have the following initial conditions: $y(0) = 6$ and $y^{\prime}(0) = 4$.
2. Solve the initial value problem, $y^{\prime\prime\prime} â€“ 6y^{\prime \prime} +11y^{\prime} – 6y = 0$, with following initial conditions:
\begin{aligned} y(0)&= 4\\ y^{\prime}(0) &= -2\\y^{\prime \prime}(0) &= -4\end{aligned}
3. Find the general solution of the differential equation, $2y^{(5)} â€“ 30y^{(4)} + 168y^{\prime\prime\prime} = 440y^{\prime\prime} â€“ 550y^{\prime} + 250y $.
Answer Key
1.
General Solution: $y = C_1e^{3x} + C_2e^{4x}$
Particular Solution: $y = 20e^{3x} â€“ 14e^{4x}$
2.$y = 15e^x â€“ 16e^{2x} + 5e^{3x}$
3. $y = C_1e^x + C_2e^{5x} + C_3xe^{5x} +C_4e^{2x}\cos x+ C_5e^{2x}\sin x$