First Order Linear Differential Equations

You might like to read well-nigh Differential Equations
and Separation of Variables outset!

A Differential Equation is an equation with a function and ane or more of its derivatives:

y + dy/dx = 5x
Example: an equation with the function y and its derivative dy dx

Here nosotros volition await at solving a special class of Differential Equations chosen First Lodge Linear Differential Equations

First Order

They are "Start Order" when there is only dy dx , not diiy dxii or diiiy dx3 etc

Linear

A showtime order differential equation is linear when it can exist fabricated to look like this:

dy dx + P(ten)y = Q(10)

Where P(x) and Q(x) are functions of ten.

To solve information technology there is a special method:

  • We invent two new functions of x, call them u and v, and say that y=uv.
  • We and then solve to find u, and so detect v, and tidy upwards and nosotros are done!

And nosotros too 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-pace method for solving them:

  • ane. Substitute y = uv, and

    dy dx = u dv dx + 5 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 tin can exist solved in the next step)
  • 4. Solve using separation of variables to notice u
  • 5. Substitute u back into the equation we got at step 2
  • half-dozen. Solve that to find 5
  • vii. Finally, substitute u and 5 into y = uv to get our solution!

Let'due south try an instance to see:

Example 1: Solve this:

dy dx − y x = one

First, is this linear? Aye, every bit it is in the form

dy dx + P(ten)y = Q(x)
where P(x) = − 1 x and Q(x) = one

So let's follow the steps:

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

So this: dy dx − y ten = 1

Becomes this: u dv dx + five du dx − uv ten = 1

Footstep ii: Factor the parts involving v

Factor five: u dv dx + v( du dx − u x ) = 1

Step three: Put the 5 term equal to zero

v term equal to nix: 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 10

Integrate: ln(u) = ln(ten) + C

Make C = ln(g): ln(u) = ln(x) + ln(k)

And so: u = kx

Stride 5: Substitute u dorsum into the equation at Pace 2

(Think v term equals 0 and so can be ignored): kx dv dx = i

Step 6: Solve this to find v

Separate variables: thousand 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 then: kv = ln(cx)

And so: v = ane 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 overnice family unit 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 whatever of those curves
the slope minus y 10 equals 1

Let's check a few points on the c=0.6 curve:

differential equation graph and points

Estmating off the graph (to one decimal place):

Point 10 y Slope ( dy dx ) dy dx − y x
A 0.6 −0.6 0 0 − −0.6 0.half-dozen = 0 + 1 = 1
B ane.six 0 1 1 − 0 1.vi = i − 0 = 1
C 2.5 1 one.4 1.four − 1 2.5 = 1.4 − 0.4 = 1

Why not exam a few points yourself? You tin can plot the curve here.

Perchance another example to help you? Maybe a picayune harder?

Instance 2: Solve this:

dy dx − 3y ten = ten

First, is this linear? Yes, as it is in the form

dy dx + P(x)y = Q(x)
where P(x) = − 3 10 and Q(x) = 10

And then let's follow the steps:

Step 1: Substitute y = uv, and dy dx = u dv dx + 5 du dx

So this: dy dx − 3y 10 = x

Becomes this:  u dv dx + v du dx − 3uv x = ten

Pace 2: Factor the parts involving v

Factor v: u dv dx + v( du dx − 3u 10 ) = ten

Footstep 3: Put the v term equal to nix

five term = zero: du dx − 3u ten = 0

So: du dx = 3u x

Step iv: Solve using separation of variables to notice u

Divide variables: du u = 3 dx x

Put integral sign: ∫ du u = three ∫ dx ten

Integrate: ln(u) = 3 ln(x) + C

Make C = −ln(thousand): ln(u) + ln(k) = 3ln(ten)

Then: u.k. = teniii

And so: u = x3 k

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

(Recollect five term equals 0 then can be ignored): ( xthree k ) dv dx = x

Footstep 6: Solve this to discover 5

Carve up variables: dv = grand x-ii dx

Put integral sign: ∫dv = ∫k x-2 dx

Integrate: v = −one thousand x-one + D

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

y = uv: y = x3 k ( −m ten-1 + D )

Simplify: y = −102 + D k ten3

Replace D/k with a single constant c: y = c x3 − x2

And it produces this squeamish family of curves:

differential equation at 0.2, 0.4, 0.6 and 0.8
y = c x3 − 102 for various values of c

And 1 more example, this time even harder:

Case 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(ten) = 2x and Q(x) = −2xiii

So let'south follow the steps:

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

And so this: dy dx + 2xy= −2x3

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

Step two: Factor the parts involving v

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

Step iii: Put the five term equal to zero

v term = zero: du dx + 2xu = 0

Stride iv: 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) = −xii + C

Make C = −ln(k): ln(u) + ln(thousand) = −x2

Then: united kingdom of great britain and northern ireland = eastward-102

And then: u = e-xii k

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

(Remember v term equals 0 so can be ignored): ( eastward-ten2 k ) dv dx = −2xiii

Step 6: Solve this to notice v

Separate variables: dv = −2k 10three eastwardx2 dx

Put integral sign: ∫dv = ∫−2k xthree ex2 dx

Integrate: v = oh no! this is hard!

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

∫RS dx = R∫Southward dx − ∫R' ( ∫South dx ) dx

(Side Note: we use R and South here, using u and v could be confusing as they already hateful something else.)

Choosing R and S is very important, this is the best option nosotros institute:

  • R = −ten2 and
  • S = 2x e102

So permit'due south go:

First pull out k: five = k∫−2x3 eastxtwo dx

R = −102 and S = 2x east102 : v = k∫(−xtwo)(2xe102 ) dx

Now integrate by parts: v = kR∫Southward dx − yard∫R' ( ∫ S dx) dx

Put in R = −102 and S = 2x eastxtwo

And likewise R' = −2x and ∫ S dx = eastwardx2

Then information technology becomes: five = −kxii ∫2x ex2 dx − yard∫−2x (eastten2 ) dx

Now Integrate: v = −kx2 eastwardten2 + k eastx2 + D

Simplify: v = kex2 (i−102) + D

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

y = uv: y = e-x2 grand ( kexii (1−10two) + D )

Simplify: y =1 − 102 + ( D m )e- xtwo

Supervene upon D/k with a single constant c: y = i − tentwo + c east- 102

And we get this nice family of curves:

differential equation
y = 1 − ten2 + c east- tentwo for various values of c

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