DIFFERENTIAL EQUATIONS 03

Initial Value Problems

An initial value problem combines a differential equation with information about the state of the system at a particular starting point.

The main idea

A differential equation often has a family of solutions. An initial condition selects the particular solution that passes through a specified starting point.

\[ \boxed{ \text{differential equation} + \text{initial condition} = \text{initial value problem} } \]

1. A family of solutions

Consider

\[ \frac{dy}{dx}=2x. \]

Integrating gives

\[ y=x^2+C. \]

Because \(C\) can take any constant value, this describes infinitely many curves.

2. Add an initial condition

Suppose we also know that

\[ y(1)=4. \]

This means that when

\[ x=1, \]

the value of \(y\) is

\[ 4. \]

Substitute into the general solution:

\[ 4=1^2+C. \]

Therefore,

\[ C=3. \]

So the particular solution is

\[ \boxed{ y=x^2+3 } \]

3. What the notation means

The notation

\[ y(x_0)=y_0 \]

means that the solution passes through the point

\[ (x_0,y_0). \]

For example,

\[ y(1)=4 \]

means that the solution passes through

\[ (1,4). \]

4. General form of a first-order initial value problem

A first-order initial value problem may be written as

\[ \boxed{ \frac{dy}{dx}=F(x,y), \qquad y(x_0)=y_0. } \]

The differential equation gives the rule of change. The initial condition gives the starting state.

5. Why the initial condition matters

The differential equation

\[ \frac{dy}{dx}=2x \]

alone does not determine a unique curve.

For example,

\[ y=x^2, \] \[ y=x^2+2, \]

and

\[ y=x^2-5 \]

all satisfy the same differential equation.

The initial condition selects one of them.

6. Visual interpretation

The differential equation gives the whole family. The initial condition identifies the required member of that family.

7. Example: exponential growth

Consider

\[ \frac{dP}{dt}=kP. \]

Its general nonzero solution is

\[ P(t)=Ae^{kt}. \]

Suppose

\[ P(0)=P_0. \]

Then

\[ P_0=Ae^0=A. \]

Therefore,

\[ \boxed{ P(t)=P_0e^{kt} } \]

8. Why \(P_0\) has a clear meaning

The notation

\[ P_0 \]

usually represents the initial value of the population.

Indeed,

\[ P(0)=P_0. \]

So the constant in the solution is determined directly by the starting population.

9. Example with separation of variables

Solve

\[ \frac{dy}{dx}=xy, \qquad y(0)=2. \]

For \(y\neq0\), separate:

\[ \frac1y\,dy=x\,dx. \]

Integrate:

\[ \ln|y|=\frac{x^2}{2}+C. \]

Exponentiate:

\[ y=Ae^{x^2/2}. \]

Use the initial condition:

\[ 2=Ae^0. \]

Therefore,

\[ A=2. \]

\[ \boxed{ y=2e^{x^2/2} } \]

10. Check the solution

A good habit is to verify both parts of the initial value problem.

First, differentiate:

\[ y=2e^{x^2/2}. \]

Then

\[ \frac{dy}{dx} = 2e^{x^2/2}x. \]

Since

\[ y=2e^{x^2/2}, \]

we have

\[ \frac{dy}{dx}=xy. \]

The differential equation is satisfied.

Also,

\[ y(0)=2. \]

So the initial condition is satisfied.

11. Initial conditions need not occur at zero

The starting point can be any specified value.

For example,

\[ y(3)=7 \]

is also an initial condition.

The word “initial” refers to the specified starting point for the problem, not necessarily to \(x=0\).

12. Second-order initial value problems

A second-order differential equation generally requires two initial conditions.

For example,

\[ \frac{d^2y}{dt^2}=-y \]

might be accompanied by

\[ y(0)=1 \]

and

\[ y'(0)=0. \]

The first condition specifies the initial position. The second specifies the initial rate of change.

13. Why more conditions are needed for higher-order equations

Integrating a first-order equation usually introduces one arbitrary constant.

Integrating a second-order equation typically introduces two.

So, in general, an \(n\)-th order differential equation requires enough independent conditions to determine the \(n\) constants in its general solution.

14. Existence is not automatically uniqueness

An initial value problem asks for a solution passing through a specified initial state.

In many well-behaved equations, this solution is unique. But this is not true for every differential equation.

So there are two separate mathematical questions:

\[ \boxed{ \text{Does a solution exist?} } \] \[ \boxed{ \text{If it exists, is it unique?} } \]

Later theory gives conditions under which both answers are yes.

15. Why initial value problems matter in modelling

A model needs both a law of change and a starting state.

For example, in a population model,

\[ \frac{dP}{dt}=F(P,t) \]

describes how the population changes.

But to predict a particular trajectory, we also need something like

\[ P(0)=P_0. \]

\[ \boxed{ \text{model dynamics} + \text{starting state} = \text{specific trajectory} } \]

16. The practical method

\[ \boxed{ 1.\; \text{Solve the differential equation generally} } \] \[ \boxed{ 2.\; \text{Keep the arbitrary constant} } \] \[ \boxed{ 3.\; \text{Substitute the initial condition} } \] \[ \boxed{ 4.\; \text{Determine the constant} } \] \[ \boxed{ 5.\; \text{Write the particular solution} } \] \[ \boxed{ 6.\; \text{Verify the equation and initial condition} } \]

17. The central idea

\[ \boxed{ \text{differential equation} \rightarrow \text{family of solutions} } \] \[ \boxed{ \text{initial condition} \rightarrow \text{one particular solution} } \]

This is the basic structure of an initial value problem.