# 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

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

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

Done!