Particular solutions using initial conditions

Solving a differential equation by integration gives a general solution: a formula with an arbitrary constant still in it.

$y = Ae^{x^2}$

Each value of $A$ gives a different curve, and every one of those curves solves the same equation. The general solution describes a whole family of curves.

A real problem always comes with one extra fact: the value of $y$ at one particular point.

That fact is called an initial condition. Cambridge uses this name even when the known value is not at $x = 0$ or $t = 0$.

Substituting that one known point fixes the value of the constant and leaves a single curve. That curve is the particular solution.

1. Find the general solution first: separate the variables and integrate.
2. Put the given values of $x$ and $y$ into the general solution.
3. Solve that equation for the constant. It is one equation in one unknown.
4. Write the solution again with the number in place of the constant.
5. Make $y$ the subject if the question asks for $y$ in terms of $x$.
6. Check: put the point back into your answer and see that both sides match.

Why one point is enough. Each value of $A$ stretches (or, if negative, flips) the same basic shape. Through any point $(x_0, y_0)$ exactly one of these curves passes, because $y_0 = Ae^{x_0^2}$ gives the single value $A = y_0e^{-x_0^2}$. So one point picks out one curve.