# 1. Solving Differential Equations (DEs)

A differential equation (or "DE") contains derivatives or differentials.

Our task is to solve the differential equation. This will involve integration at some point, and we'll (mostly) end up with an expression along the lines of "y = ...".

Recall from the Differential section in the Integration chapter, that a differential can be thought of as a derivative where dy/dx is actually not written in fraction form.

### Examples of Differentials

dx (this means "an infinitely small change in x")

d\theta (this means "an infinitely small change in \theta")

dt (this means "an infinitely small change in t")

## Examples of Differential Equations

### Example 1

We saw the following example in the Introduction to this chapter. It involves a derivative, dy/dx:

(dy)/(dx)=x^2-3

As we did before, we will integrate it. This will be a general solution (involving K, a constant of integration).

So we proceed as follows:

y=int(x^2-3)dx

and this gives

y=x^3/3-3x+K

But where did that dy go from the (dy)/(dx)? Why did it seem to disappear?

In this example, we appear to be integrating the x part only (on the right), but in fact we have integrated with respect to y as well (on the left). DEs are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time.

We could have written our question only using differentials:

dy = (x2 − 3)dx

(All I did was to multiply both sides of the original dy/dx in the question by dx.)

Now we integrate both sides, the left side with respect to y (that's why we use "dy") and the right side with respect to x (that's why we use "dx") :

int dy = int(x^2 - 3)dx

Then the answer is the same as before, but this time we have arrived at it considering the dy part more carefully:

y=x^3/3-3x+K

On the left hand side, we have integrated int dy = int 1 dy to give us y.

Note about the constant: We have integrated both sides, but there's a constant of integration on the right side only. What happened to the one on the left? The answer is quite straightforward. We do actually get a constant on both sides, but we can combine them into one constant (K) which we write on the right hand side.

### Example 2

This example also involves differentials:

\theta^2 d\theta = sin(t + 0.2)\ dt

We have:

A function of theta with d theta on the left side, and

A function of t with dt on the right side.

To solve this, we would integrate both sides, one at a time, as follows:

int theta^2 d theta = int sin(t+0.2)dt

{theta^3}/3 = -cos(t + 0.2) + K

We have integrated with respect to θ on the left and with respect to t on the right.

Here is the graph of our solution, taking K=2:

Continues below

## Solving a differential equation

From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps.

It is important to be able to identify the type of DE we are dealing with before we attempt to solve it.

## Definitions

First order DE: Contains only first derivatives

Second order DE: Contains second derivatives (and possibly first derivatives also)

Degree: The highest power of the highest derivative which occurs in the DE.

### Example 3 - Order and Degree

a) (d^2y)/(dx^2)+((dy)/(dx))^3-3x+2y=8

This DE has order 2 (the highest derivative appearing is the second derivative) and degree 1 (the power of the highest derivative is 1.)

b) ((dy)/(dx))^5-2x=3\ sin(x)-sin(y)

This DE has order 1 (the highest derivative appearing is the first derivative) and degree 5 (the power of the highest derivative is 5.)

c) (y'')^4+2(y')^7-5y=3

This DE has order 2 (the highest derivative appearing is the second derivative) and degree 4 (the power of the highest derivative is 4.)

## General and Particular Solutions

When we first performed integrations, we obtained a general solution (involving a constant, K).

We obtained a particular solution by substituting known values for x and y. These known conditions are called boundary conditions (or initial conditions).

It is the same concept when solving differential equations - find general solution first, then substitute given numbers to find particular solutions.

Let's see some examples of first order, first degree DEs.

### Example 4

a. Find the general solution for the differential equation

dy + 7x\ dx = 0

b. Find the particular solution given that y(0)=3.

Here is the graph of the particular solution we just found:

### Example 5

Find the particular solution of

y' = 5

given that when x=0, y=2.

### Example 6

Find the particular solution of

y''' = 0

given that:

y(0) = 3, y'(1) = 4, y''(2) = 6

### Example 7

After solving the differential equation,

(dy)/(dx)ln\ x-y/x=0

(we will see how to solve this DE in the next section Separation of Variables), we obtain the result

y=c ln x

Did we get the correct general solution?

## Second Order DEs

We include two more examples here to give you an idea of second order DEs. We will see later in this chapter how to solve such Second Order Linear DEs.

### Example 8

The general solution of the second order DE

y'' + a2y = 0

is

y = A cos ax + B sin ax

### Example 9

The general solution of the second order DE

y'' − 3y' + 2y = 0

is

y = Ae2x + Bex

If we have the following boundary conditions:

y(0) = 4, y'(0) = 5

then the particular solution is given by:

y = e2x + 3ex

Now we do some examples using second order DEs where we are given a final answer and we need to check if it is the correct solution.

### Example 10 - Second Order DE

Show that

y = c_1 sin 2x + 3 cos 2x

is a general solution for the differential equation

(d^2y)/(dx^2)+4y=0

### Example 11 - Second Order DE

Show that (d^2y)/(dx^2)=2(dy)/(dx) has a solution of y = c1 + c2e2x

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