# Question #0c06b

Jan 15, 2018

See below.

#### Explanation:

Solve this differential equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} - 1 = 0$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\to \int \mathrm{dy} = \int \mathrm{dx}$

Solving for the general solution gives:

$y = x + C$

Where $C$ is the constant of integration. As $C$ is yet undetermined and can take any (complex) value we call this the general solution. $C$ might be $+ 2$ or $- 3$ (though without further information there is no reason it can't be something like $1 , 4 , \pi , \sqrt{10}$ or $1 + i$). These two values of $C$ would give us:

$y = x + 2$ or $y = x - 3$

In these cases $C$ has been determined so these are particular solutions. In other words they are just specific cases of the general solution. The value of $C$ can usually be worked out when given "initial" or "boundary" conditions.

Jan 15, 2018

For $y = x - 3$ we have $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$ so $\frac{\mathrm{dy}}{\mathrm{dx}} - 1 = 0$
And also, for for $y = x + 2$ we have $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$ so $\frac{\mathrm{dy}}{\mathrm{dx}} - 1 = 0$