What do the phrases "once upon a time", "in the beginning", and "on a dark and stormy night" all have in common? They all indicate the start of a story! That is what the initial value for a differential equation tells you, where to start. But unlike the fact that two stories that start with "once upon a time" aren't the same, all solutions to initial value problems might not be so unique. So read through to figure out how to make sure you get a unique solution to your problem!
Definition of an Initial-Value Problem
Let's start by answering a common question. Is it supposed to be "initial-value" or "initial value"? Well, it depends on what country you are in and who wrote the textbook you are reading! To be consistent this article doesn't use the hyphen. Instead initial value problem is abbreviated IVP, which gets around the question of whether that hyphen should be there or not.
So what is an IVP?
An IVP is a differential equation together with a place for a solution to start. They are often written
\[ \begin{align} &y' = f(x,y) \\ &y(a) = b \end{align} \]
where \((a,b)\) is the point the solution \(y(x)\) must go through.
Remember that the first step to solving a differential equation is to try and find the general solution, which gives you a constant of integration. Then using the initial value, you can find a particular solution. For more information on these topics see General Solutions to Differential Equations and Particular Solutions to Differential Equations.
First Order Differential Equation Initial Value Problem
Let's start with the constant coefficient first order linear differential equation
\[ y'+Ay=B\]
where \(A\) and \(B\) are real numbers with \(A \neq 0\). Remember that the general solution to this linear differential equation is
\[y(x) =Ce^{-Ax}+\frac{B}{A}\]
where \(C\) is a constant.
If you add the initial value \(y(x_1) = y_1\) where \(x_1\) and \(y_1\) are real numbers, then you can plug those into the general solution to get
\[ y_1=Ce^{-Ax_1}+\frac{B}{A}, \]
so
\[ C = \frac{y_1 - \frac{B}{A}}{e^{-Ax_1}}. \]
That means the solution to the IVP
\[ \begin{align} y' &= f(x,y) \\ y(x_1) &= y_1 \end{align} \]
is
\[ \begin{align} y(x) &= \frac{y_1 - \frac{B}{A}}{e^{-Ax_1}} e^{-Ax}+\frac{B}{A} \\ &= \left(y_1 - \frac{B}{A} \right)e^{-A(x-x_1)} +\frac{B}{A} . \end{align}\]
That can be a bit of a pain to memorize, so rather than doing that it is best to remember the general solution and then solve given your initial values. Let's take a look at an example.
Solve the IVP
\[ \begin{align} &y' +3y = 7 \\ &y(0) = \frac{1}{3}. \end{align} \]
Solution
The first step is finding the general solution, which for this problem is
\[y(x) =Ce^{-3x}+\frac{7}{3}.\]
Remember that this is actually a family of functions, and you can graph them for various values of \(C\).
The general solution is a family of functions.
The IVP is asking you to pick out the function that goes through the point \(\left(0, \frac{1}{3}\right)\). Using the initial value to solve for \(C\), you get
\[\frac{1}{3} =Ce^{-3\cdot 0}+\frac{7}{3},\]
so \(C = -2\). That means the solution to the IVP is
\[y(x) = -2e^{-3x}+\frac{7}{3}.\]
It is one particular function out of the family of functions that has the property that it goes through the initial value, as you can see in the graph below.
The particular solution goes through the initial value.
Solutions to linear constant coefficient first order differential equation IVPs have some nice properties that all come directly from the fact that you can explicitly write down the solution as
\[ y(x) = \left(y_1 - \frac{B}{A} \right)e^{-A(x-x_1)} +\frac{B}{A}.\]
Those properties are,
If you have two different initial values, the solutions corresponding to those initial values cannot cross. If they could cross it would violate the uniqueness part.
Solutions exist on the whole real line.
Solutions all have the same long term behavior. In other words, for a solution \(y(x)\) of the IVP, \[ \lim\limits_{x \to \infty} y(x) =\begin{cases} \frac{B}{A} & \mbox{ if } A>0 \\ \infty & \mbox{ if } A<0\quad \mbox{provided that } y_1-\frac{B}{A}>0 \\ -\infty & \mbox{ if } A<0\quad \mbox{provided that } y_1-\frac{B}{A}<0\end{cases}. \]
Now let's move on to the first order linear differential equation
\[ y'+P(x)y=Q(x)\]
where \(P(x)\) and \(Q(x)\) are functions. To find the solution to this differential equation you use the integrating factor
\[ \alpha(x)=e^{\int P(x)\,\mathrm{d}x},\]
and the general solution is
\[ y(x) = \frac{1}{\alpha (x)} \left( \int \alpha(x) \, Q(x) \, \mathrm{d}x + C \right).\]
Finding the solution to the IVP
\[ \begin{align} y'+P(x)y&=Q(x) \\ y(a) &= b \end{align} \]
isn't going to be nearly as nice as in the constant coefficient case since you need to do the integration before you can figure out what \(C\) is. So let's look at an example.
Solve the IVP
\[ \begin{align} &y'+\frac{1}{x}y=\frac{1}{\sqrt{x}} \\ &y(1) = \frac{5}{3} .\end{align} \]
Solution
First let's find the general solution. The integrating factor is
\[ \begin{align} \alpha(x)&=e^{\int P(x)\,\mathrm{d}x} \\ &= \exp\left( \int \frac{1}{x} \,\mathrm{d}x \right) \\ &= |x|. \end{align} \]
Notice that the initial value has \(x=1\), so in other words you are interested in positive values for \(x\) and you can drop the absolute value in the integrating factor to write
\[ \alpha(x) = x.\]
Then the general solution is given by
\[ \begin{align} y(x) &= \frac{1}{\alpha (x)} \left( \int \alpha(x) \, Q(x) \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \int x \frac{1}{\sqrt{x}} \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \int \sqrt{x} \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \frac{2}{3}x^{\frac{3}{2} }+ C \right) \\ &= \frac{2\sqrt{x}}{3} + \frac{C}{x}. \end{align}\]
Now you can plug in the initial value \( y(1) = \dfrac{5}{3} \) to get
\[ \frac{5}{3} = \frac{2\sqrt{1}}{3} + \frac{C}{1}, \]
so
\[C = \frac{5}{3} - \frac{2}{3} = 1.\]
With the assumption that \(x>0\), the solution to the IVP is
\[ y(x) = \frac{2\sqrt{x}}{3} + \frac{1}{x}.\]
Notice that there was only one solution, but it certainly doesn't exist on the whole real line! For examples of a first order linear IVP without a solution, or with infinitely many solutions, take a look at Particular Solutions to Differential Equations.
You might wonder when a first order linear IVP will have a unique solution that exists on the whole real line.
If \(a, b \in \mathbb{R}\), and \(P(x)\), \(Q(x)\) are both continuous functions on the whole real line 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 on the whole real line.
Initial value Problems and Separable Differential Equations
Remember that a differential equation is separable if you can write it in the form
\[N(y)y' = M(x)\]
where \(N(y)\) and \(M(x)\) are functions. For more information and examples of this kind of equation see Separable Equations.
To make this into an IVP, all you need to do is pick an initial value. So for real numbers \(a\) and \(b\), the IVP is
\[\begin{align} &N(y)y' = M(x) \\ &y(a) = b. \end{align}\]
With separable equations, you often need to be careful with the interval of existence for solutions. In cases like that, the initial value tells you which of the intervals to choose as the interval of existence.
Let's take a look at an example.
If possible, solve the IVP
\[ \begin{align} &y' = \frac{y^2}{x} \\ &y(0) = 4. \end{align} \]
Solution
First, you can rewrite the differential equation as
\[ \frac{1}{y^2} y' = \frac{1}{x}, \]
so it is a separable differential equation. Separating variables and integrating gives you
\[ \int \frac{1}{y^2} \, \mathrm{d}y = \int \frac{1}{x} \, \mathrm{d}x\]
so
\[ -\frac{1}{y} = \ln |x| + C.\]
Now let's try and use the initial conditions. If you do, you will be using \(x=0\), and you can't take the natural log of zero. So in fact this IVP has no solution.
One more example, to see the kinds of things that can happen.
Consider the IVP
\[ \begin{align} &y' = y^{\frac{1}{3}} \\ &y(0) = 0. \end{align} \]
First show that the constant solution \(y(x)=0\) satisfies this IVP. Then see if there are any other solutions.
Solution
Certainly if you take the derivative of a constant function you get zero, and \(0^\frac{1}{3} = 0\), so the constant function \(y(x) = 0\) satisfies the differential equation. It also satisfies the initial condition, so it does satisfy the IVP.
What about other solutions? This is a separable equation, so separating and integrating gives you
\[ \int \frac{1}{y^\frac{1}{3} } \, \mathrm{d}y = \int 1\, \mathrm{d}x\]
so
\[ \frac{3}{2}y^\frac{2}{3} = x+ C.\]
Using the initial value \(y(0) = 0\) to find \(C\), you get
\[ \frac{3}{2}0^\frac{2}{3} = 0+ C,\]
so \(C=0\). That means there is a second solution to this IVP, namely the implicit solution
\[ \frac{3}{2}y^\frac{2}{3} = x.\]
You can get the explicit solution by solving for \(y\). If you do, you get
\[ y^\frac{2}{3} = \frac{2}{3}x,\]
so
\[ y(x) =\left( \frac{2}{3}x \right)^{\frac{3}{2}}.\]
Notice that the maximal interval of existence is \( [0,\infty) \) since you can't take the square root of a negative number. If you graph the two functions, you can see that both the constant function and the function \(y(x)\) that you solved for both satisfy the equation and the initial value.
Two solutions to an IVP
More examples are always good!
Examples of Differential Equation Initial Value Problems
Let's look more examples involving IVPs.
Find the solution to the IVP
\[ \begin{align} &2xy'+4y=3 \\ &y(2) = \frac{5}{4} \end{align} \]
by first showing that the general solution is
\[y(x) = \frac{C}{x^2} + \frac{3}{4}.\]
What is the interval of existence for the solution to the IVP?
Solution
To see that \(y(x)\) is a general solution, you will need to find the derivative and plug it into the equation. The derivative is
\[ y'(x) = \frac{-2C}{x^3} .\]
Plugging that in, you get
\[ \begin{align} 2xy'+4y &= 2x\left(\frac{-2C}{x^3} \right) + 4\left(\frac{C}{x^2} + \frac{3}{4} \right) \\ &= \frac{-4C}{x^2} + \frac{4C}{x^2} + 4\left(\frac{3}{4} \right) \\ &= 3, \end{align}\]
which is the right hand side of the original differential equation. Therefore \(y(x)\) is a general solution.
Notice that the general solution is not defined at \(x = 0\). Therefore the interval of existence is either going to be \( (-\infty , 0)\) or \((0, \infty)\). Since the initial value is when \(x=2\), the interval of existence for the IVP is \((0, \infty)\).
Now to solve the IVP. Using the initial value,
\[ y(2) = \frac{C}{2^2} + \frac{3}{4} = \frac{5}{4}. \]
Solving for \(C\), you can see that \(C=2\). So the solution to the IVP is
\[y(x) = \frac{2}{x^2} + \frac{3}{4}.\]
Sometimes you will have a solution which is implicit, and that can be used to solve an IVP as well.
Is
\[ y^2 = x^2 - 3\]
an implicit solution to the IVP
\[ \begin{align} &y' = \frac{x}{y} \\ &y(2) = 1 ?\end{align} \]
Solution
This requires the use of implicit differentiation. But before doing that, it is a good idea to make sure the proposed solution satisfies the initial value, because if it doesn't then it certainly can't be a solution to the IVP! Checking, the left hand side gives you \((y(2))^2 = 1^2 = 1\), and the right hand side gives you \(2^2 - 3 = 1\). Since they are the same, the proposed implicit solution at least satisfies the initial value.
Then using implicit differentiation and remembering that \(y\) is a function of \(x\),
\[ \frac{\mathrm{d}}{\mathrm{d} x} (y(x))^2 = \frac{\mathrm{d}}{\mathrm{d} x} \left(x^2 - 3 \right).\]
Looking first at the left hand side, and using the fact that if it is a solution then
\[y' = \frac{x}{y} ,\]
you get
\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x} (y(x))^2 &=2y(x)y'(x) \\ &= 2y\frac{x}{y} \\ &= 2x.\end{align}\]
On the right hand side,
\[ \frac{\mathrm{d}}{\mathrm{d} x} \left(x^2 - 3 \right) = 2x.\]
Since both sides are the same, \(y(x)\) does satisfy the differential equation.
Therefore the proposed implicit solution does actually solve the IVP.
What do you do if you can't find an implicit or explicit solution to an IVP?
Numerical Initial Value Problems in Ordinary Differential Equations
Most IVPs you have seen here you could find an explicit solution for, or at least an implicit one! What do you do if you have an IVP that you can't solve like that? That is when you need to use numerical methods to find solutions to IVPs. The method that most people start with is Euler's Method. It uses an initial value, and the differential equation, to calculate slopes that lead to an approximation of the solution to an ordinary differential equation IVP. For more information on this topic, along with plenty of examples, see Euler's Method.
Initial Value Problem Differential Equations - Key takeaways
- In differential equations, initial value problem is often abbreviated IVP.
- An IVP is a differential equation together with a place for a solution to start, called the initial value.
- IVPs are often written\[ \begin{align} &y' = f(x,y) \\ &y(a) = b \end{align} \] where \((a,b)\) is the point the solution \(y(x)\) must go through.
- When solving separable IVPs it is important to choose the interval of existence that contains the initial value.
- Not all IVPs have a unique solution, or even have a solution!
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