### Topics

## Introduction to Differential Equations - At A Glance:

Differential equations have two kinds of solutions: **general** and **particular**.

The **general solution** to a differential equation is the collection of all solutions to that differential equation. A general solution will usually contain some undetermined constants.

### Sample Problem

*y* = 4*x* + *C* is the general solution to the d.e.

*y *' = 4.

### Sample Problem

*y* = 3*x*^{2} + *Bx* + *C* is the general solution to the d.e.

*y *" = 6.

A **particular solution** to a differential equation is a solution with all the constants filled in.

### Sample Problem

The function *y* = 4*x* + 2 is a specific solution to the d.e.

*y *' = 4.

Initial value problems usually have a particular solution only, because the initial condition(s) force us to pick values for the constant(s) in the general solution.

#### Example 1

Find a particular solution to the IVP *y*' = 4 and *y*(1) = 9.
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We know any solution to the d.e. *y*' = 4
has the form *y* = 4*x* + *C*.
If we're to have *y*(1) = 9, then 9 = 4(1) + *C* and so *C* has to equal 5. There's no choice. This means the specific solution to this IVP is *y* = 4*x* + 5.
The usual process for solving IVPs is to - get a general solution by solving the d.e., then
- get a particular solution by using the initial condition(s) to find the constants.
We'll be able to solve more kinds of differential equations later, but we can do some practice ones now by thinking backwards. | |

#### Example 2

Find the particular solution to the IVP *y*' = 6*x* and *y*(0) = 3.
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The general solution to the d.e. *y*' = 6*x* is *y* = 3*x*^{2} + *C*
Convince yourself: take the derivative of *y* = 3*x*^{2} + *C*. For the particular solution, we want to have *y* = 3 when *x* = 0, so we can plug these numbers into the general solution and solve for *C*: 3 = 3(0)^{2} + *C* so *C* = 3. The particular solution to this IVP is *y* = 3*x*^{2} + 3.
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#### Example 3

Solve the IVP *y*" = 12*x* and *y*'(0) = 5 and *y*(0) = 4
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We need to think backwards twice to find the general solution. Thinking backwards from *y*" to *y*', we get *y*' = 6*x*^{2} + *B*.
Thinking backwards again, we see that *y* must look like this: *y* = 2*x*^{3} + B*x* + *C*.
To find the particular solution, let's first use the condition *y*(0) = 4. Plugging *x* = 0 and *y* = 4 into our equation for *y* gives us 4 = 2(0)^{3} + *B*(0) + *C* so *C* = 4. To find *B* we take the condition *y*'(0) = 5 and plug *x* = 0 and *y*' = 5 into the equation for *y*': 5 = 6(0)^{2} + *B* so *B* = 5. The particular solution to this IVP is *y* = 2*x*^{3} + 5*x* + 4.
In the previous example we had a second-order differential equation (link to defn of second-order) and two initial conditions. For a first-order differential equation we need only one initial condition. A differential equation of order *n* will have a general solution with n undetermined constants, and you need n initial conditions to figure out a particular solution. | |

#### Example 4

Solve the IVP *f* ^{(3)}(*x*) = 0, *f* (0) = 0, *f* '(0) = 1, *f* "(0) = 4.
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First we find the general solution. If the third derivative of *f* is 0, the second derivative of *f* must be a constant: *f* ^{(2)}(*x*) = *A*.
The first derivative must be a linear function with slope A: *f *'(*x*) = *Ax* + *B*.
The original function must look like this: Now we use the initial conditions to fill in the constants. When *x* = 0 we want f(0) = 0, so which means *C* = 0. When *x* = 0 we want *f* ' (0) = 1, so 1 = *A*(0) + *B* which means *B* = 1. When *x* = 0 we want *f* "(0) = 4, so 4 = *A*. Putting everything we know about the constants together, we get Of course, we can check our answer by taking 3 derivatives of *f* to make sure that *f* ^{(3)}(*x*) = 0 (it does) and making sure that *f*, *f* ', and *f* " have the correct values when *x* = 0. | |

#### Exercise 1

Solve the initial value problem.

*y*' = 4*x*^{3}, *y*(2) = 20

Answer

First we find the general solution. If *y*' = 4*x*^{3} then

*y* = *x*^{4} + *C*.

Since we're given the initial condition that *y* = 20 when *x* = 2, we have

20 = (2)^{4} + *C* = 16 + *C*.

This means *C* = 4 and so the particular solution is

*y* = *x*^{4} + 4.

#### Exercise 2

Solve the initial value problem.

*y*' = -sin *x*, *y*(0) = 2

Answer

The general solution is

*y* = cos *x* + *C*.

Using the initial condition *y*(0) = 2, we have

2 = cos 0 + *C* = 1 + *C*

so *C* = 1. The particular solution is

*y* = cos *x* + 1.

#### Exercise 3

Solve the initial value problem.

*y*' = *x*^{2} + *x*^{3} + 2, *y*(1) = 3

Answer

The general solution is

Using the initial condition that says *y* = 3 when *x* = 1, we have

Solving, we get and so the particular solution is

#### Exercise 4

Solve the initial value problem.

*y*" = 4*x*, *y*(0) = 7, *y*'(0) = 5

Answer

We need to think backwards twice. The first time, we get

*y*' = 2*x*^{2} + *B*.

The second time, we get

When *x* = 0 we want *y* to be 7, which means

so *C* = 7. When *x* = 0 we want *y*' to be 5, so we get

5 = 2(0)^{2} + *B*

and so *B* = 5. The particular solution is

#### Exercise 5

Solve the initial value problem.

*y*' = -*e*^{x}, *y*(0) = -3

Answer

The general solution is

*y* = -*e*^{x} + *C*.

Using the initial condition, we have

-3 = -*e*^{0} + *C* = -1 + *C*.

We conclude that *C* = -2 and so the particular solution is

*y* = -*e*^{x} - 2.