**An Introduction**

Simply put, a differential equation is an equation that involves an unknown function and it's derivatives. Your goal is to solve for the function. Students usually take a course in differential equations after finishing calculus.

Example of differential equations include:

(I). dy/dx = 2*x + 5

(II). 2y' - y = 0

(III). dy/dx + dy/dt = y - t

(IV). (sin x) * dx/dt + (cos y) * dx/dy = 1

Ordinary differential equations involve unknown functions of only one variable. From above, equations (I) and (II) are ordinary differential equations. Partial differential equations involve unknown multi-variable functions. Examples include equations (III) and (IV).

In this series, I will concentrate on ordinary differential equations.

The order of a differential equation is determine by the largest-order derivative. All four equations, (I) through (IV), are differential equations of the first-order.

An example of a second-order differential equation is

y'' + 2*y' - 3*y = e^x.

We are going to be begin with the most basic differential equation, dy/dx = f(x). **Solving dy/dx = f(x)**

When solving equations in the form dy/dx = f(x), it is matter of knowing your integral calculus, because:

dy/dx = f(x)

All that needs to be done is to take the integral of both sides and we get:

y(x) = ∫ f(x) dx + C

where C is some constant. We'll talk about C later.

Need to brush up on your anti-derivatives? Here are some common ones:

∫ a dx = a*x

∫ x^n dx = x^(n+1)/(n+1)

∫ e^x dx = e^x

∫ ln x dx = 1/x

∫ sin x dx = -cos x

∫ cos x dx = sin x

Integration by Parts:

∫ u'(x)*v(x) dx = u(x)*v(x) - ∫ u(x)*v'(x) dx

There are far more complete tables in reference books or online.

Now let's get to work.

1. dy/dx = 2*x + 4

Simple as it gets, just take the integral of 2x + 4 and we get:

y = x^2 + 4*x + C

2. y' = 2*e^(2*x)

We have:

y = ∫ 2*e^(2*x) dx + C = e^(2*x) + C

**Initial Conditions**

Differential equations may have additional conditions called initial conditions. Initial conditions give the value of the function (and it's derivatives - to order n-1) at some given point. Initial conditions given at x=0 is common. Initial conditions allow us to solve for the constant (C).

3. dy/dx = 2*x + 4, y(1)=0

Here we have an initial condition y(1)=0. Any time we have initial conditions, we solve the differential equation first, then solve for the constant.*

So:

dy/dx = 2*x + 4

y = x^2 + 4*x + C

With y(1)=0, substitute x=1 and y=0 and solve for C.

0 = 1^2 + 4*1 + C

C = -5

Our final answer is y = x^2 + 4*x - 5.

4. y' = 2*e^(2*x), y(0)=2

y' = 2*e^(2*x)

y = e^(2*x) + C

Taking in the initial condition y(0)=2:

2 = e^(2*0) + C

C = 1

Our final answer is y = e^(2*x) + 1.

* The procedure is slightly different when it comes to using Laplace Transforms, which is later on in the series.

We have come to the conclusion of Part 1. Next time in this series, we will be talking about separable equations. Hope to see you then! Eddie

This blog is property of Edward Shore. 2013

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