First Order Linear Differential Equations

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:

y + dy/dx = 5x
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

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:

differential equation at 0.2, 0.4, 0.6, 0.8 and 1.0
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:

differential equation graph and points

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:

differential equation at 0.2, 0.4, 0.6 and 0.8
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:

differential equation
y = 1 − x2 + c e- x2 for various values of c

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