**Contents:**

What is an Initial Condition?

Differential Equation Initial Value Problem Example.

## What is an Initial Condition?

An **initial condition **is a starting point; Specifically, it gives dependent variable values (or one of its derivatives) for a certain independent variable. It allows you to zoom in on a specific solution.

In general, an initial condition can be *any* starting point. For example, you might want to define an initial pressure or a starting balance in a bank account. In statistics, it’s a nuisance parameter in unit root testing (Muller & Elliot, 2003). In calculus, the term usually refers to the starting condition for finding the particular solution for a differential equation.

## Initial Condition at a Discontinuity

Any smooth function will have one initial condition. Sometimes, you might have a discontinuous function, which is broken up into parts. For simplicity’s sake, lets say your function is broken up into two parts. You’ll have a “first” condition (sometimes called the pre-initial condition)*before*the discontinuity and a “second” condition (sometimes called the post-initial condition)

*after*the discontinuity.

## Use in Differential Equations

The “initial” condition in a differential equation is usually what is happening when the **initial time (t) **is at zero (Larson & Edwards, 2008). For example, let’s say you have some function g(t), you might be given the following initial condition:

- g(0) = 40 (the function returns a value of 40 at t = 0 seconds)
- g
^{′(0)}= 32 (the value of the first derivative at t = 0 seconds is 32)

An initial condition leads to a particular solution; If you don’t have an initial value, you’ll get a general solution.

## Differential Equation Initial Value Problem.

Watch the video for two examples:

Can’t see the video? Click here. When a differential equation specifies an initial condition, the equation is called an

**initial value problem**. Initial conditions require you to search for a particular (specific) solution for a differential equation.

For example, the differential equation needs a general solution of a function or series of functions (a general solution has a constant “c” at the end of the equation):

^{dy}⁄_{dx} = 19x^{2} + 10

But if an initial condition is specified, then you must find a particular solution (a single function). For example:

^{dy}⁄_{dx}19x^{2} + 10; **y(10) = 5.**

Finding a particular solution for a differential equation requires one more step—simple substitution—*after *you’ve found the general solution.

## Initial Value Problem: Examples

**Example Problem 1: **Solve the following differential equation, with the initial condition y(0) = 2.

^{dy}⁄_{dx} = 10 – x

Step 1: **Use algebra **to move the “dx” to the right side of the equation (this makes the equation more familiar to integrate):

^{dy}⁄_{dx} = 10 – x →

dy = 10 – x dx

Step 2: **Integrate both sides of the equation**.

- ∫dy = ∫10 – x dx →
- ∫1 dy = ∫10 – x dx →
- y = 10x –
^{x2}⁄_{2}+ C

Step 3: **Substitute in the values specified in the initial condition. **In this sample problem, the initial condition is that when x is 0, y=2, so:

- 2 = 10(0) –
^{02}⁄_{2}+ C - 2 = 0 + C
- C = 2

Therefore, the function that satisfies this particular differential equation with the initial condition y(0) = 2 is y = 10x – ^{x2}⁄_{2} + 2

*That’s it! *

**Initial Value Example problem #2: **Solve the following initial value problem: ^{dy}⁄_{dx} = 9x^{2} – 4x + 5; y(-1) = 0

Step 1: **Rewrite the equation, using algebra**, to make integration possible (essentially you’re just moving the “dx”.

^{dy}⁄_{dx}= 9x^{2}– 4x + 5 →- dy = (9x
^{2}– 4x + 5) dx

Step 2: **Integrate both sides of the differential equation** to find the general solution:

- ∫ dy = ∫(9x
^{2}– 4x + 5) dx → - ∫ 1 dy = ∫(9x
^{2}-4x + 5) dx → - y = (
^{9x3}⁄_{3}–^{4x2}⁄_{2}+ 5x + C → - y=3x
^{3}-2x^{2}+ 5x + C

Step 3: **Evaluate the equation** you found in Step 3 for when **x = -1** and **y = 0**.

0 = 3(-1)^{3} -2(-1)^{2} + 5(-1) + C →

0 = -3 -2 – 5 + C →

0 = -10 + C

c = 0

Therefore, the particular solution to the initial value problem is y = 3x^{3} – 2x^{2} + 5x + 10

*That’s it!*

## References

Larson, R. & Edwards, B. (2008). Calculus of a Single Variable. Cengage Learning.

MIT Open Courseware. Initial Conditions. Retrieved July 19, 2020 from: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iii-fourier-series-and-laplace-transform/unit-step-and-unit-impulse-response/MIT18_03SCF11_s25_1text.pdf

Muller, U. & Elliot, G. (2003). Tests for Unit Roots. Econometrica, Vol. 71, No. 4 (July), 1269–1286

Calculus.

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