Q.Particular solution of differential equation $\dfrac{dy}{dx}+x=0$ at $x=0, y=1$ will be -

Initial Value Problem

The intuition: a rule of change plus a starting point

A car's speed at time $t$ is $v(t)=\dfrac{ds}{dt}=2t$. Can you say where the car is at $t=5$? Not yet — you don't know where it started ($0$ m? $10$ m? $100$ m?). The differential equation gives the rule of change, but you also need one starting snapshot to pin down the actual motion. Supply "$s=5$ when $t=0$" and now everything is determined.

That pairing — a differential equation together with an initial condition — is an Initial Value Problem (IVP).

The precise statement

An IVP has two parts:

1. A differential equation, e.g. first-order: $\dfrac{dy}{dt}=f(t,y)$.
2. An initial condition, the value at a starting point: $y(t_0)=y_0$.

Written together,

and the goal is the particular function $y(t)$ satisfying both.

A worked example

Solve $\dfrac{dy}{dt}=3y,\ \ y(0)=2$.

First solve the equation, ignoring the condition. Separating and integrating, $\dfrac{dy}{y}=3\,dt$ gives $\log|y|=3t+C$, so the general solution is $y=Ae^{3t}$. Now apply $y(0)=2$: $\,2=Ae^{0}=A$. Hence the unique solution is

$y(t)=2e^{3t}.$ …