In antiderivatives with initial conditions you solved equations like $\frac{dy}{dx} = 3x^{2}$, $y(0) = 5$. A differential equation is the general version: an equation that relates an unknown function to its own derivatives, such as $y' = 2y$ (growth in proportion to size) or $y'' = -9y$ (a spring). The unknown is a whole function, not a number.
Why it works
A solution must make the equation true for every $x$ in an interval, so you check it by substituting the whole function and its derivatives and simplifying until both sides match identically. For $y' = 2y$, try $y = Ce^{2x}$: then $y' = 2Ce^{2x} = 2y$ for every $x$ and every constant $C$. So one equation has a whole family of solutions. An initial condition such as $y(0) = 3$ picks one member: $C = 3$.
Key idea
A differential equation relates a function and its derivatives. Its order is the highest derivative that appears. A solution is a function that satisfies the equation identically on an interval. The general solution is the family of all solutions (usually with constants); an initial value problem adds conditions that pick one.
Worked example: Checking a solution
Show that $y = 3e^{2x}$ solves $y' = 2y$.
- Differentiate: $y' = 6e^{2x}$.
- Right side: $2y = 2\cdot 3e^{2x} = 6e^{2x}$.
- The two sides are equal for every $x$, so it is a solution.
Answer: It satisfies the equation for all $x$.
Common mistake
Checking at a single point. Plugging in $x = 0$ only shows that the equation holds there; the function $y = 3 + 6x$ also has $y(0) = 3$ and $y'(0) = 6 = 2y(0)$, yet it is not a solution, since $y' = 6$ while $2y = 6 + 12x$. Check: after substituting, simplify both sides as expressions in $x$ and make sure they agree everywhere.
Worked example: Exponential guesses
For which values of $r$ is $y = e^{rx}$ a solution of $y'' - y' - 6y = 0$?
- $y' = re^{rx}$ and $y'' = r^{2}e^{rx}$.
- Substitute: $(r^{2} - r - 6)e^{rx} = 0$. Since $e^{rx}$ is never 0, $r^{2} - r - 6 = 0$.
- $(r - 3)(r + 2) = 0$, so $r = 3$ or $r = -2$.
Answer: $r = 3$ and $r = -2$
Worked example: An initial value problem
Every solution of $y' = 4 - y$ has the form $y = Ce^{-x} + 4$. Find the solution with $y(0) = 1$.
- Check the family: $y' = -Ce^{-x}$ and $4 - y = -Ce^{-x}$. They match.
- At $x = 0$: $1 = C + 4$, so $C = -3$.
- $y = 4 - 3e^{-x}$.
Answer: $y = 4 - 3e^{-x}$
Constant functions are worth checking first. For $y' = y(3 - y)$ the constants $y = 0$ and $y = 3$ make the right side 0, and their derivative is 0, so they are solutions. These equilibrium solutions come back in the logistic equation.