Initial value problem

An initial value problem (IVP) is a Differential equation together with initial conditions that specify the function and its derivatives at a single point. The DE alone has a family of solutions; the initial conditions pin down a specific one.

For an $n$-th order ODE, you need $n$ initial conditions: the value of the function and the values of its first $n-1$ derivatives, all at the same point $x_0$:

$y(x_0)=y_0,y^{\prime }(x_0)=y_0^{\prime },\dots ,y^{(n-1)}(x_0)=y_0^{(n-1)}$

Why $n$ conditions for an $n$-th order ODE

The general solution of an $n$-th order ODE has $n$ arbitrary constants. To pin them all down uniquely, you need $n$ equations — one per condition.

For first-order: $\frac{dy}{dx}=3x^2$ has general solution $y=x^3+C$. One arbitrary constant, one initial condition needed: $y(0)=5\Rightarrow C=5$.

For second-order: $y^{\prime \prime }=-y$ has general solution $y=c_1\cos x+c_2\sin x$. Two constants, so two conditions: typically $y(x_0)$ (position) and $y^{\prime }(x_0)$ (velocity).

Example

The IVP $\frac{dy}{dx}=3x^2$, $y(0)=5$ has the unique solution $y(x)=x^3+5$.

For a second-order example: $y^{\prime \prime }=y-x$, $y(0)=1$, $y^{\prime }(0)=0$ pins down a specific solution from the family $y=c_1{e}^x+c_2{e}^{-x}+x$.

Initial vs boundary conditions

Two kinds of “side conditions” for ODEs:

• Initial conditions: $y(x_0)=y_0$, $y^{\prime }(x_0)=y_0^{\prime }$, etc. — all at the same point. Used in IVPs.
• Boundary conditions: function values specified at different points, e.g., $y(0)=0$, $y(1)=1$. Used in boundary value problems (BVPs), which are common for PDEs (and rare for ODEs).

For ODEs, almost everything is an IVP. BVPs become important in PDEs (heat equation, wave equation) where you specify behavior on the boundary of a region.

When does the IVP have a solution?

The Existence and uniqueness theorem gives sufficient conditions: if $f(x,y)$ in $y^{\prime }=f(x,y)$ is continuous near $(x_0,y_0)$ and $\frac{\partial f}{\partial y}$ is also continuous, then a unique local solution exists.

If both conditions hold, the solution is guaranteed and unique in some interval around $x_0$. If they fail, the IVP might have multiple solutions, no solution, or solutions that exist only briefly.

Solving IVPs

The general procedure:

1. Find the general solution of the DE (using whichever method fits — separation of variables, integrating factor, characteristic equation, Laplace transform, etc.).
2. Apply the initial conditions to solve for the arbitrary constants.
3. Substitute back to get the particular solution.

For example, $y^{\prime \prime }+y=0$, $y(0)=1$, $y^{\prime }(0)=2$:

• General solution: $y=c_1\cos x+c_2\sin x$.
• $y(0)=c_1=1$, so $c_1=1$.
• $y^{\prime }(x)=-c_1\sin x+c_2\cos x$, so $y^{\prime }(0)=c_2=2$.
• Particular solution: $y(x)=\cos x+2\sin x$.

For solving IVPs efficiently, especially with discontinuous forcing terms, the Laplace transform is the most powerful tool.

For the case where the relationship between $x$ and $y$ is given implicitly, see Implicit solution to ODE.