# First Order Linear 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 d2y dx2 or d3y dx3 etc

## 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

into

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

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 curves
the 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.

### 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

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 = x3
And so:u = x3 k

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

(Remember v term equals 0 so can be ignored):( x3 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 = x3 k ( −k x-1 + D )
Simplify:y = −x2 + D k x3
Replace D/k with a single constant c: y = c x3 − x2

And it produces this nice family of curves: y = c x3 − x2
for various values of c

And one more example, this time even harder:

### Example 3: Solve this:

dy dx + 2xy= −2x3

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) = −2x3

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

So this: dy dx + 2xy= −2x3
Becomes this: u dv dx + v du dx + 2xuv = −2x3

Step 2: Factor the parts involving v

Factor v:u dv dx + v( du dx + 2xu ) = −2x3

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 = −2x dx
Integrate:ln(u) = −x2 + C
Make C = −ln(k):ln(u) + ln(k) = −x2
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 = −2x3

Step 6: Solve this to find v

Separate variables:dv = −2k x3 ex2 dx
Put integral sign: dv = −2k x3 ex2 dx
Integrate:v = oh no! this is hard!

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

RS dx = RS 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 = −x2 and
• S = 2x ex2

So let's go:

First pull out k:v = k−2x3 ex2 dx
R = −x2 and S = 2x ex2:v = k(−x2)(2xex2) dx
Now integrate by parts:v = kRS dx − kR' ( S dx) dx

Put in R = −x2 and S = 2x ex2

And also R' = −2x and S dx = ex2

So it becomes:v = −kx22x ex2 dx − k−2x (ex2) dx
Now Integrate:v = −kx2 ex2 + k ex2 + D
Simplify:v = kex2 (1−x2) + D

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

y = uv:y = e-x2 k ( kex2 (1−x2) + D )
Simplify:y =1 − x2 + ( D k )e-x2
Replace D/k with a single constant c: y = 1 − x2 + c e-x2

And we get this nice family of curves: y = 1 − x2 + c e-x2
for various values of c

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