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 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
intody 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 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.
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 = 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
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= −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 = −2∫x 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 = 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 = −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 = kR∫S dx − k∫R' ( ∫ S dx) dx
Put in R = −x2 and S = 2x ex2
And also R' = −2x and ∫ S dx = ex2
So it becomes:v = −kx2∫2x 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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Differential Equations Separation of Variables Calculus Index
