Generally, you like to eat lunch every day, but what time do you eat it? Do you prefer before noon, noon, or after noon to eat? The specific time you like to eat lunch is a particular solution to the general question of when you like to eat. You can do the same thing with differential equations. A general solution has a constant in it, but a particular solution to a differential equation doesn't.
What Is the Difference Between the General and Particular Solution of a Differential Equation?
A general solution to a differential equation is one that has a constant in it. It is really a family of functions that solves the differential equation.
A particular solution to a differential equation is one that satisfies an initial value.
In other words, you are able to pick one particular solution from the family of functions that solves the differential equation, but also has the additional property that it goes through the initial value.
A linear first-order differential equation can be written as
\[ y' + P(x)y = Q(x)\]
where \(P(x)\) and \(Q(x)\) are functions. You can see how to find solutions to this type of differential equation in the article Linear Differential Equations. These solutions have a constant of integration in them and make up a family of functions that solve the equation.
If you add an initial value to the linear first order differential equation you get what is called an initial value problem (often written IVP). It will look like
\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]
where \(P(x)\) and \(Q(x)\) are functions, and \(a\) and \(b\) are real-valued constants. Because you have an initial value, the solution to this initial value problem is exactly one function, not a family of them. It is a particular solution to the more general linear first-order differential equation without an initial value.
Finding a Particular Solution to Linear Differential Equation
Let's look at an example to see how you would find a particular solution to a linear differential equation.
Consider the linear differential equation initial value problem
\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(1) = 7 .\end{align}\]
First, find the general solution, then find the particular solution if possible.
Solution:
First, let's solve the differential equation to get the general solution. Here \(P(x) = -1/x\) and \(Q(x) = 3x\), so you know the integrating factor is
\[ \begin{align} \exp\left( -\int \frac{1}{x} \, \mathrm{d} x\right) &= \exp\left(-\log x\right) = \frac{1}{x}.\end{align} \]
That means the solution to
\[ y' -\frac{y}{x} = 3x \]
is given by
\[ \begin{align} y\left(\frac{1}{x}\right) &= \int 3x\left(\frac{1}{x}\right)\, \mathrm{d}x \\ &= \int 3 \, \mathrm{d}x \\ &= 3x + C. \end{align}\]
Then solving for \(y\) you get
\[ y(x) = 3x^2 + Cx.\]
So the general solution is \(y(x) = 3x^2 + Cx \).
The particular solution makes use of the initial values to figure out what \(C\) is. Here the initial value is \(y(1) = 7\). Plugging that into the general solution you get
\[ 7 = 3(1)^2 + C\cdot 1,\]
or
\[ 4 = C.\]
So the particular solution to the initial value problem is
\[ y(x) = 3x^2 + 4x.\]
Not all first-order linear initial value problems have a solution.
Let's go back to the linear differential equation, but with a different initial value. Is there a particular solution to
\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(0) = 7 \end{align}\]
Solution:
From the previous example, you know that the general solution to
\[ y' -\frac{y}{x} = 3x \]
is
\[ y(x) = 3x^2 + Cx.\]
Now try plugging in the initial value to find \(C\). When you do,
you get
\[ 7 = 3(0)^2 + C\cdot 0,\]
or
\[ 7 = 0.\]
Hey, wait a minute! Seven doesn't equal zero, so what gives? Since you can't find a \(C\) that satisfies the initial value, this initial value problem doesn't have a particular solution!
Sometimes you even get more than one solution!
Let's go back to the linear differential equation, but with a different initial value. Is there a particular solution to
\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(0) = 0 \end{align}\]
Solution:
From the previous example you know that the general solution to
\[ y' -\frac{y}{x} = 3x \]
is
\[ y(x) = 3x^2 + Cx.\]
Now try plugging in the initial value to find \(C\). When you do,
you get
\[ 0 = 3(0)^2 + C\cdot 0,\]
or
\[ 0= 0.\]
Hey, wait a minute, that is always true! It doesn't matter what value of \(C\) you put in, it will always satisfy the initial value. That means this initial value problem has infinitely many solutions!
So why does this happen? It turns out that the existence of a solution, and the uniqueness of a solution, depend on the functions \(P(x)\) and \(Q(x)\).
If \(a, b \in \mathbb{R}\), and \(P(x)\), \(Q(x)\) are both continuous functions on the interval \((x_1, x_2)\) where \(x_1 < a < x_2 \) then the solution to the initial value problem
\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]
exists and is unique.
In other words, the difficulty with the differential equation
\[ y' -\frac{y}{x} = 3x \]
is that the function
\[ P(x) = -\frac{1}{x} \]
is not a continuous function at \(x=0\), so any initial value that goes through \(x=0\) may not have a solution, or may not have a unique solution.
Particular Solutions to Nonhomogeneous Differential Equations
First, recall that a homogeneous first-order linear differential equation looks like
\[ y' + P(x)y = 0.\]
But that is just a special case of the first-order linear differential equation you have already seen! In other words, the first order linear nonhomogeneous differential equation looks like
\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]
where \(P(x)\) and \(Q(x)\) are functions, and \(a\) and \(b\) are real-valued constants. So all you need to do to find more information on these kinds of equations is to look at the article Nonhomogeneous Linear Equations.
Particular Solutions to Separable Differential Equations
A first-order separable differential equation is an equation that can be written in the form
\[y'=f(x)g(y).\]
For more information on these types of differential equations, you can take a look at our articles Separable Equations and Application of Separation of Variables.
Just like with first-order linear differential equations, you get a family of functions as the solution to separable equations, and this is called a general solution. On the other hand, the solution to the initial value problem
\[\begin{align} &y'=f(x)g(y) \\ &y(a)=b \end{align}\]
is a particular solution.
Let's take a look at an example.
Find the particular solution to the initial value problem
\[ \begin{align} & y' = \dfrac{y^2}{x} \\ & y(1) = 2 \end{align}\]
along with any domain restrictions it might have.
Solution:
First let's find the solution. Separate the variables to get
\[ \frac{1}{y^2} y' = \frac{1}{x} \]
and then integrate both sides with respect to \(x\) to get
\[ \int \frac{1}{y^2} \, \mathrm{d} y = \int \frac{1}{x} \, \mathrm{d} x \]
so
\[ -\frac{1}{y} = \ln |x| + C.\]
Then solving for \(y\), the general solution is given by
\[ y(x) = -\frac{1}{ \ln |x| + C }.\]
Now you can use the initial condition \(y(1)=2\) to find a particular solution. That means
\[ 2 = -\frac{1}{ \ln |1| + C },\]
and
\[C = -\frac{1}{2}.\]
So the particular solution is
\[ y(x) = -\frac{1}{ \ln |x| - \frac{1}{2} }.\]
Now let's look at any restrictions that might be on the solution. With the absolute value signs there, you don't need to worry about taking the log of a negative number. However you still can't have \(x=0\), and you also need that the denominator is not zero. That means you need
\[ \ln |x| - \frac{1}{2} \ne 0.\]
Using properties of logarithms, you can see that \(x \ne \pm \sqrt{e}\) is also a necessary condition.
That means there are four intervals that your solution might be in:
- \( -\infty < x < -\sqrt{e} \)
- \( -\sqrt{e} < x < 0 \)
- \(0 < x < \sqrt{e}\)
- \( \sqrt{e} < x < \infty\).
So how do you know which one your solution is in? Just look at the initial value! The initial value for this problem is \(y(1) = 2 \), and \(x=1\) is in the interval \( (0 , \sqrt{e} )\). That means the domain restriction for this particular solution is \( (0 , \sqrt{e} )\).
Examples of a Particular Solution to a Differential Equation
Let's look at some examples of particular solutions. First, how do you know if something is really a particular solution?
Show that
\[ y = 2x^{-3}\]
is a particular solution of the initial value problem
\[ \begin{align} &xy' +3y = 0 \\ &y(1) = 2. \end{align}\]
Solution:
It is usually a good idea to check the initial value first since it will be relatively easy, and if the prospect doesn't satisfy the initial value it can't be a solution to the initial value problem. In this case,
\[ \begin{align} y(1) & = 2(1)^{-3} \\ &= 2, \end{align}\]
so the function \(y(x) = 2x^{-3} \) does satisfy the initial value. Now you just need to check to see if it satisfies the equation. For that you need \(y'\), so
\[ y' = 2(-3)(x^{-4}) = -6x^{-4}.\]
Substituting that into the differential equation,
\[ \begin{align} xy' +3y &= x\left(-6x^{-4} \right) + 3\left(2x^{-3} \right) \\ &= -6x^{-3} + 6x^{-3} \\ &= 0 \end{align}\]
So the proposed solution does satisfy the differential equation.
Since \(y(x) = 2x^{-3} \) satisfies both the initial value and the differential equation, it is a particular solution to the initial value problem.
Let's take a look at something that isn't first order.
Find a particular solution to the initial value problem
\[ \begin{align} &y'' = 3x+2 \\ &y(0)=3 \\ &y'(0) = 1. \end{align}\]
Solution:
The first step is to find a general solution. Notice that this is actually a second-order equation, so it has two initial values. However this is an especially nice second-order equation since the only \(y\) in it is a second derivative, and it is already separated.
Integrating both sides of the equation with respect to \(x\) you get
\[ y' = \frac{3}{2}x^2 + 2x + C.\]
Integrating once more you get
\[ y(x) = \frac{1}{2}x^3 + x^2 + Cx + D,\]
which is the general solution. There are two constants to go with the two initial values. Using \(y'(0) = 1 \) you get
\[ y'(0) = \frac{3}{2}0^2 + 2(0) + C = 1,\]
So \(C = 1\). Plugging that in to the general solution gives you
\[ y(x) = \frac{1}{2}x^3 + x^2 + x + D,\] and then you can use the second initial value \(y(0)=3 \) to get
\[ y(0) = \frac{1}{2}0^3 + 0^2 +0 + D = 3,\]
which means that \(D = 3\). Therefore the particular solution to the initial value problem is
\[ y(x) = \frac{1}{2}x^3 + x^2 + x + 3.\]
Particular Solutions to Differential Equations - Key takeaways
- The first-order linear equation \[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]
where \(P(x)\) and \(Q(x)\) are functions, and \(a\) and \(b\) are real-valued constants is called an initial value problem.
The solution to an initial value problem is called a particular solution.
The solution to a differential equation without initial values is called a general solution. It is a family of functions rather than a single particular one.
The solution to the first order separable initial value problem
\[\begin{align} &y'=f(x)g(y) \\ &y(a)=b \end{align}\]
is a particular solution.
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