# What does it mean for a function to be a solution of a differential equation?

What does it mean for a function to be a solution of a differential equation? I think I understand, but I still don't have a simple or intuitive understanding.

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What does it mean for a number to be a solution of an equation? – Neal Dec 9 '11 at 3:43
A number is a solution when it fits within the range of answers that the function generates. For differential equations an equation is a solution when the equation is created satisfies the derivative's range. – user17366 Dec 9 '11 at 3:53

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Not quite. An equation is a statement about numbers involving an unknown. A number solves an equation if, when substituted for the unknown, it makes the statement true. Likewise, a differential equation is a statement about functions involving an unknown function. A function solves a differential equation if, when substituted, the statement is true. – Neal Dec 9 '11 at 12:16
Simple? If you "plug in" the function to the differential equation and it gives an equality, then it's a solution! – The Chaz 2.0 Dec 14 '11 at 2:28

## 2 Answers

Suppose that $\Omega \subset \mathbb{R}^{n+1}$ and that you have some function $F: \Omega \to \mathbb{R}$. Then $F$ defines the differential equation $$F(x, y, y', ..., y^{(n-1)}) = 0$$ A solution to this differential equation is a function $y: (a,b) \to \mathbb{R}$ defined on some open interval (which could be all of $\mathbb{R}$) such that $$(x, y(x), y'(x), ..., y^{(n-1)}(x)) \in \Omega$$ for all $x \in (a,b)$ and $$F(x, y(x), y'(x), ..., y^{(n-1)}(x)) = 0$$ for all $x \in (a,b)$. Since every ordinary differential equation can be written in such a way this defines the notion of solution unambiguously. Intuitively, you can think of this as "$F$ being $0$ on the graph of $y$ and its derivatives." In practice you usually add some requirements on $F$ and $\Omega$, for example it is usual to require that $\Omega$ is a region, and that $F$ is continuous and invertible around $0$.

To give an example, if you have the equation $$y''(x) = 2x^3y(x) + y'(x)^2$$ then you can take $$F(x_1, x_2, x_3, x_4) = 2x_1^3x_2 + x_3^2 - x_4$$

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Solving $x+1=3$ means finding a value for $x$ that satisfies the equation $x+1=3$. In this case, $x=2$ does the job because $2+1=3$.

Now, given a differential equation such as:

$\frac{dy}{dx}= 2x$ ..........(1)

a solution to this equation is a function (call it $y(x)$). Of course not any function will do. The correct function must satisfy the equation we have in (1) above.

for this particular case, y(x) can be equal to $x^2$.

Why? Because the derivative of $x^2$ with respect to $x$ is $\frac{dy}{dx}= 2x$ which is what we have in our equation.

But wait, what about $y(x)=x^2$+5?

In fact, this is another solution. As you can see, in this case we end up with many solutions all of the form:

$y(x)=x^2+k$

where k is a constant. This is because, all of such functions satisfy our differential equation.

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