Solution of 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:


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   + P(x)y = Q(x)
dx

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   =  u dv   +  v du
dx dx dx

Steps

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

  • 1. Substitute y = uv, and
    dy   =  u dv   +  v du
    dx dx dx
    into
    dy   + P(x)y = Q(x)
    dx
  • 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: 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 = 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

Those curves are interesting: they are the solution to the equation   dy dx y x = 1

In other words:

Anywhere on any of those curves
it will be true that 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 learn? Maybe a little harder?

Example: 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 a hard one!

Example: 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 would just be confusing as they are already mean something else here.)

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 )ex2
Replace D/k with a single constant c:   y = 1−x2 + c ex2

Done!