# Can Integration constant be anything?

When the question is to solve a given ODE (without initial value), can I assign any value to the constant $C$, in order to solve it, or is there a specific constant value that must be found?

In other words can two people assume different combination to the constant C to solve ODE, or one of them would be wrong?

Example: $$\frac{dy}{y}=\frac{xdx}{x^2+1}$$ $$\ln(y)=\ln\sqrt{x^2+1}+C$$ So in order to solve it I assume $C=\ln(e^c)$, but why? Can a different person assume different value? or is this the only value in the world that must be assumed for $C$ in order to solve it? Solution:$$y=C\sqrt{x^2+1}$$

• Edits and improvements are welcome Commented Jul 15, 2016 at 0:50
• There are some restrictions on $C$. For instance, it cannot be a banana. Commented Jul 15, 2016 at 0:53
• We done that just to get rid off $\ln$ from the solution $\ln y=\ln(\sqrt{x^2+1})+C$. You have seen your self that by letting $C=\ln e^c$ we have reached such a simple form of the solution as $y=C\sqrt{x^2+1}$, which is simple and elegant. Now in the final solution, in place of $C$ you can put any value as long as it represents a constant. Commented Jul 15, 2016 at 0:57
• lol =)) hahaha you funny Commented Jul 15, 2016 at 0:57
• @IttayWeiss Why not? The derivative of a constant banana is still zero. ðŸ˜œ Commented Jul 15, 2016 at 0:59

In general, a differential equation problem consists of two parts: a system of equations and boundary conditions.

Suppose we're dealing with functions of time. The system of equations describes how quantities change over time. The boundary conditions—if they are provided—tell you about how quantities take on specific values at specific times. In this way, the boundary conditions constrain the space of solutions.

Here's a simple example from calculus: I tell you that $$y^\prime = 2x.$$ This is a system containing one equation. There are many solutions to this equation: $y(x) = x^2$, for example, or $y(x) = x^2 - 3$, or $y(x) = x^2 + 17$. Notice (by plugging in) that each of these different functions satisfies the differential equation. In general, for shorthand, we write that the set of solutions is just

$$y(x) = x^2 + C \qquad (\text{for any }C)$$ where each new value of $C$ gives you a different solution, and every solution to the equation can be expressed this way.

But in addition to providing the differential equation $y^\prime = 2x$, I might also give you a boundary condition such as $y(2) = 7$.

Now we must only consider solutions to the differential equation that satisfy this new constraint. In particular, out of the infinitely many possible solutions of the form $y(x) = x^2 + C$, only one of them satisfies the boundary condition— namely,

$$y_\star(x) = 2x + 3.$$

Differential equations may have many— even infinitely many — solutions. Some of these solutions will depend on a choice of constant. The general rule is that any function you find which satisfies the differential equation is a valid solution.

Boundary conditions impose additional constraints— if you have boundary conditions, the solution must satisfy not only the differential equation, but the boundary conditions at well.

• Got it. Thank you for the help! Commented Jul 15, 2016 at 14:34

The differential equation has a solution that is unique up to multiplication by the constant $C$. In your case $C$ must be in the range of $\ln e^{c}$ with $c \in \mathbb{R}$ so $C$ can be any real number. This means you have an infinite set of solutions. If I pick something different, say my solution is $y=g(x)$ as long as there is some constant $C$ you can pick so $g(x)=C\sqrt{x^{2}+1}$ we're both okay; our functions are basically the same but mine is yours stretched or compressed vertically.

When you solve an ODE like this you're coming up with a set of solutions. In order to get a specific one you usually need an initial value for the function which will restrict your choice of $C$ to a single, distinct constant, e.g., $y(x_0)=y_0$ where $x_0$ and $y_0$ are just numbers.

• Thank you for the answer :) Commented Jul 15, 2016 at 14:34