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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
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

3 Answers 3

When you write an algebraic equation you are trying to generalise a calculation for say. In a way you are saying what will be the end result if you give certain input or vice-versa.

Similarly when you write a function you are generalising the notion of the end result itself. After defining a function you can opt for any result your wish and find the input required to get that desired result.

In the same way when you define a differential equation you are actually trying to emphasis that in what way the rate of change of a function and its value will merge together to give the result. This kind of approach is useful in study of complex systems where the rate of changes can be measured with time and which helps in understanding the nature of system mathematically.

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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:


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

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