You might like to read about Differential Equations

and Separation of Variables first!

A Differential Equation is an equation with a function and one or more of its derivatives:

Example: an equation with the function **y** and its derivative** dy dx **

Here we will look at solving a special class of Differential Equations called **First Order Linear Differential Equations**

## First Order

They are "First Order" when there is only ** dy dx ** , not

**or**

*d*^{2}y**dx**^{2}**etc**

*d*^{3}y**dx**^{3}## Linear

A **first order differential equation** is **linear** when it can be made to look like this:

*dy* **dx** + P(x)y = Q(x)

Where **P(x)** and **Q(x)** are functions of x.

To solve it there is a special method:

- We invent two new functions of x, call them
**u**and**v**, and say that**y=uv**. - We then solve to find
**u**, and then find**v**, and tidy up and we are done!

And we also use the derivative of **y=uv** (see Derivative Rules (Product Rule) ):

*dy* **dx** = u *dv* **dx** + v *du* **dx**

## Steps

Here is a step-by-step method for solving them:

- 1. Substitute
**y = uv**, and*dy***dx**= u*dv***dx**+ v*du***dx***dy***dx**+ P(x)y = Q(x) - 2. Factor the parts involving
**v** - 3. Put the
**v**term equal to zero (this gives a differential equation in**u**and**x**which can be solved in the next step) - 4. Solve using separation of variables to find
**u** - 5. Substitute
**u**back into the equation we got at step 2 - 6. Solve that to find
**v** - 7. Finally, substitute
**u**and**v**into**y = uv**to get our solution!

Let's try an example to see:

Example 1: Solve this:

*dy* **dx** − *y* **x** = 1

First, is this linear? Yes, as it is in the form

*dy* **dx** + P(x)y = Q(x)

where **P(x) = − 1 x ** and

**Q(x) = 1**

So let's follow the steps:

Step 1: Substitute **y = uv**, and ** dy dx = u dv dx + v du dx **

So this: *dy* **dx** − *y* **x** = 1

Becomes this:u *dv* **dx** + v *du* **dx** − *uv* **x** = 1

Step 2: Factor the parts involving **v**

Factor **v**:u *dv* **dx** + v( *du* **dx** − *u* **x** ) = 1

Step 3: Put the **v** term equal to zero

**v** term equal to zero: *du* **dx** − *u* **x** = 0

So: *du* **dx** = *u* **x**

Step 4: Solve using separation of variables to find **u**

Separate variables: *du* **u** = *dx* **x**

Put integral sign: ∫ *du* **u** = ∫ *dx* **x**

Integrate:ln(u) = ln(x) + C

Make C = ln(k):ln(u) = ln(x) + ln(k)

And so:u = kx

Step 5: Substitute **u** back into the equation at Step 2

(Remember **v** term equals 0 so can be ignored):kx *dv* **dx** = 1

Step 6: Solve this to find **v**

Separate variables:k dv = *dx* **x**

Put integral sign: ∫k dv = ∫ *dx* **x**

Integrate:kv = ln(x) + C

Make C = ln(c):kv = ln(x) + ln(c)

And so:kv = ln(cx)

And so:v = *1* **k** ln(cx)

Step 7: Substitute into **y = uv** to find the solution to the original equation.

y = uv:y = kx *1* **k** ln(cx)

Simplify:y = x ln(cx)

And it produces this nice family of curves:

y = x ln(cx) for various values of **c**

What is the meaning of those curves?

They are the solution to the equation ** dy dx − y x = 1**

In other words:

**Anywhere on any of those curvesthe slope minus y x equals 1**

Let's check a few points on the **c=0.6** curve:

Estmating off the graph (to 1 decimal place):

Point | x | y | Slope ( dy dx ) | dy dx − y x |
---|---|---|---|---|

A | 0.6 | −0.6 | 0 | 0 − −0.6 0.6 = 0 + 1 = 1 |

B | 1.6 | 0 | 1 | 1 − 0 1.6 = 1 − 0 = 1 |

C | 2.5 | 1 | 1.4 | 1.4 − 1 2.5 = 1.4 − 0.4 = 1 |

Why not test a few points yourself? You can plot the curve here.

Perhaps another example to help you? Maybe a little harder?

Example 2: Solve this:

*dy* **dx** − *3y* **x** = x

First, is this linear? Yes, as it is in the form

*dy* **dx** + P(x)y = Q(x)

where **P(x) = − 3 x ** and

**Q(x) = x**

So let's follow the steps:

Step 1: Substitute **y = uv**, and ** dy dx = u dv dx + v du dx **

So this: *dy* **dx** − *3y* **x** = x

Becomes this:u *dv* **dx** + v *du* **dx** − *3uv* **x** = x

Step 2: Factor the parts involving **v**

Factor **v**:u *dv* **dx** + v( *du* **dx** − *3u* **x** ) = x

Step 3: Put the **v** term equal to zero

**v** term = zero: *du* **dx** − *3u* **x** = 0

So: *du* **dx** = *3u* **x**

Step 4: Solve using separation of variables to find **u**

Separate variables: *du* **u** = 3 *dx* **x**

Put integral sign: ∫ *du* **u** = 3 ∫ *dx* **x**

Integrate:ln(u) = 3 ln(x) + C

Make C = −ln(k):ln(u) + ln(k) = 3ln(x)

Then:uk = x^{3}

And so:u = *x ^{3}*

**k**

Step 5: Substitute **u** back into the equation at Step 2

(Remember **v** term equals 0 so can be ignored):( *x ^{3}*

**k**)

*dv*

**dx**= x

Step 6: Solve this to find **v**

Separate variables:dv = k x^{-2} dx

Put integral sign: ∫dv = ∫k x^{-2} dx

Integrate:v = −k x^{-1} + D

Step 7: Substitute into **y = uv** to find the solution to the original equation.

y = uv:y = *x ^{3}*

**k**( −k x

^{-1}+ D )

Simplify:y = −x^{2} + *D* **k** x^{3}

Replace **D/k** with a single constant **c**: y = c x^{3 }− x^{2}

And it produces this nice family of curves:

y = c x^{3 }− x^{2} for various values of **c**

And one more example, this time even ** harder**:

Example 3: Solve this:

*dy* **dx** + 2xy= −2x^{3}

First, is this linear? Yes, as it is in the form

*dy* **dx** + P(x)y = Q(x)

where **P(x) = 2x** and **Q(x) = −2x ^{3}**

So let's follow the steps:

Step 1: Substitute **y = uv**, and ** dy dx = u dv dx + v du dx **

So this: *dy* **dx** + 2xy= −2x^{3}

Becomes this:u *dv* **dx** + v *du* **dx** + 2xuv = −2x^{3}

Step 2: Factor the parts involving **v**

Factor **v**:u *dv* **dx** + v( *du* **dx** + 2xu ) = −2x^{3}

Step 3: Put the **v** term equal to zero

**v** term = zero: *du* **dx** + 2xu = 0

Step 4: Solve using separation of variables to find **u**

Separate variables: *du* **u** = −2x dx

Put integral sign: ∫ *du* **u** = −2∫x dx

Integrate:ln(u) = −x^{2} + C

Make C = −ln(k):ln(u) + ln(k) = −x^{2}

Then:uk = e^{-x2}

And so:u = *e ^{-x2}*

**k**

Step 5: Substitute **u** back into the equation at Step 2

(Remember **v** term equals 0 so can be ignored):( *e ^{-x2}*

**k**)

*dv*

**dx**= −2x

^{3}

Step 6: Solve this to find **v**

Separate variables:dv = −2k x^{3} e^{x2} dx

Put integral sign: ∫dv = ∫−2k x^{3} e^{x2} dx

Integrate:v = oh no! this is hard!

Let's see ... we can integrate by parts... which says:

∫RS dx = R∫S dx − ∫R' ( ∫S dx ) dx

*(Side Note: we use R and S here, using u and v could be confusing as they already mean something else.)*

Choosing R and S is very important, this is the best choice we found:

- R = −x
^{2}and - S = 2x e
^{x2}

So let's go:

First pull out k:v = k∫−2x^{3} e^{x2} dx

**R = −x ^{2}** and

**S = 2x e**:v = k∫(−x

^{x2}^{2})(2xe

^{x2}) dx

Now integrate by parts:v = kR∫S dx − k∫R' ( ∫ S dx) dx

Put in R = −x^{2} and S = 2x e^{x2}

And also R' = −2x and ∫ S dx = e^{x2}

So it becomes:v = −kx^{2}∫2x e^{x2} dx − k∫−2x (e^{x2}) dx

Now Integrate:v = −kx^{2} e^{x2} + k e^{x2} + D

Simplify:v = ke^{x2} (1−x^{2}) + D

Step 7: Substitute into **y = uv** to find the solution to the original equation.

y = uv:y = *e ^{-x2}*

**k**( ke

^{x2}(1−x

^{2}) + D )

Simplify:y =1 − x^{2} + ( *D* **k** )e^{-}^{x2}

Replace **D/k** with a single constant **c**: y = 1 − x^{2} + c e^{-}^{x2}

And we get this nice family of curves:

y = 1 − x^{2} + c e^{-}^{x2} for various values of **c**

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Differential Equations Separation of Variables Calculus Index