# 2. Separation of Variables

### Later, on this page...

Some differential equations can be solved by the method of
**separation of variables** (or "variables separable") . This method is only possible if
we can write the differential equation in the form

A(x)dx+B(y)dy= 0,

where *A*(*x*) is a function of *x* only and
*B*(*y*) is a function of *y* only.

Once we can write it in the above form, all we do is integrate throughout, to obtain our general solution.

**NOTE:** In this variables separable section we only deal with **first order, first
degree** differential equations.

### Example 1 - Separation of Variables form

**a)** The differential equation (which we saw earlier in Solutions of Differential Equations):

`(dy)/(dx)ln\ x-y/x=0`

**can** be expressed in the required
form, *A*(*x*) *dx* + *B*(*y*) *dy* = 0, after some algebraic juggling:

` (dy)/(dx)ln\ x-y/x=0`

`dy\ ln\ x-(y\ dx)/x=0`

`dy-(y\ dx)/(x\ ln\ x)=0`

`(dy)/y-(dx)/(x\ ln\ x)=0`

`1/ydy-1/(x\ ln\ x)dx=0`

Here, `A(x) = -1/(x\ ln\ x)` and `B(y) = 1/y`.

[To solve the equation, we would then integrate throughout].

**b)** The following differential equation **cannot** be
expressed in the required form, so it cannot be solved using
separation of variables:

`(dy)/(dx)=(3(x+y))/(x(y-2))`

### Example 2

Solve the differential equation:

y^{2}dy+x^{3}dx =0

Answer

This is already in the required form (since the x-terms are together with dx terms, and y-terms are together with dy terms), so we simply integrate:

`inty^2dy+intx^3dx=0`

Giving:

`y^3/3+x^4/4=K`

This is the **general solution** for the differential
equation.

We can continue on to solve this as an explicit function in x, as follows:

`y^3=3(K-x^4/4)`

`y=root(3)(3(K-x^4/4))`

Taking a typical constant value `K=5`, we have this solution graph:

Typical solution graph `y=root(3)(3(5-x^4/4))`.

Get the Daily Math Tweet!

IntMath on Twitter

### Example 3

Solve the differential equation: `2(dy)/(dx)=(y(x+1))/x`

Answer

First we must separate the variables:

`(2\ dy)/y=((x+1)dx)/x`

This gives us:** **`(2\ dy)/y=(1+1/x)dx`

We now integrate:

`int(2 dy)/y=int(1+1/x)dx`

`2 ln y=x+ln x+K`

Go back to Integration: Basic Logarithm Form if you are rusty on this integration.

We could continue with our solution and express `y` as an explicit function of `x`, as follows:

`ln y=(x+ln x+K)/2`

`y=e^((x+ln x+K)//2)`

Taking a typical constant value `K=1`, we have this solution graph:

Typical solution graph `y=e^((x+ln x+1)//2)`.

### Example 4

Solve `t(dx)/(dt)-x=3`

Answer

`t(dx)/(dt)-x=3`

We want all *x*'s on one side, all the terms in *t* on the other.

`t(dx)/(dt)=x+3`

Divide both sides by `(x + 3)` and multiply both sides by `dt`:

`t(dx)/(x+3)=dt`

Next, divide both sides by *t*.

`(dx)/(x+3)=(dt)/t`

Integrate both sides.

`int(dx)/(x+3)=int(dt)/t`

**Important:** *dx *and *dt* must be on **top** of the fraction (i.e. in the numerator).

`ln|x+3|=ln|t|+C`

Take "*e* to power of both sides":

`x+3=e^(ln|t|+C)=e^(ln|t|)e^C=Kt`

(We let `e^C=K`, constant)

So `x = Kt - 3`.

#### Typical solution and graph

We let `K = 7`, for illustration):

Typical solution graph `x=7t-3`.

Easy to understand math videos:

MathTutorDVD.com

### Example 5

Solve** **`sqrt(1+4x^2)\ dy=y^3x\ dx`

Answer

Separating variables gives us:

`(dy)/y^3=(x\ dx)/(sqrt(1+4x^2))`

Integrating gives us:

`int(dy)/y^3=int(x\ dx)/(sqrt(1+4x^2)`

We now proceed to integrate the 2 sides separately. That is, we integrate the left side in *y* only (since after separating the variables we have terms in *y* and a *dy* on the left) and we work on the right side in *x* only (since we have terms in *x* and a *dx* only on the right).

For the right hand side involving *x*, let *u* = (1 + 4*x*^{2}), so *du* = 8*x* *dx* and *du*/8 = *x dx.*

`int(dy)/y^3=1/8int(du)/sqrtu`

`=(-1)/(2y^2)`

`=1/8(2)u^(1"/"2)+K`

`=1/4sqrt(1+4x^2)+K`

So the solution is given by:

`(-1)/(2y^2)=1/4sqrt(1+4x^2)+K`

We could go on to solve this in *y*, as follows:

Multiply both sides by `−2`:

`1/y^2=-1/2sqrt(1+4x^2)-2K`

For convenience, introduce a new variable `K_1 = -4K`, so that we'll have `-2K=K_1/2`. Our solution becomes:

`1/y^2=-1/2sqrt(1+4x^2)+K_1/2 =` ` (K_1 - sqrt(1+4x^2))/2`

Take the reciprocal of both sides:

`y^2=2/(K_1-sqrt(1+4x^2))`

Then solve for *y*:

`y=(+-sqrt2)/sqrt(K_1-sqrt(1+4x^2))`

(The constant *K*_{1} can be chosen so that the expression in the denominator is real.)

Get the Daily Math Tweet!

IntMath on Twitter

Here is the graph of a typical solution for Example 5 where we have taken `K=50`:

Typical solution graph `y=(+-sqrt2)/sqrt(50-sqrt(1+4x^2))`.

## Particular Solutions

Our examples so far in this section have involved some constant of integration, *K*.

We now move on to see particular solutions, where we know some boundary conditions and we substitute those into our general solution to give a **particular solution**.

### Example 6

Find the particular solution for

`(dy)/(dx)+2y=6`

given that *x *= 0 when *y =* 1.

Answer

We solve as before, and then use the given information to find the value of the unknown *K.*

Separating variables:

`(dy)/(dx)+2y=6`

`(dy)/(dx)=6-2y`

`dy=(6-2y)dx`

`(dy)/(6-2y)=dx`

Integrating gives:

`int(dy)/(6-2y)=intdx`

`(-1)/2ln|6-2y|=x+K`

[For the integral involving `y`, we put `u = 6 - 2y` giving `du = - 2\ dy`. This means we'll replace `dy` with `(-1/2)du` and integrate `(1/u)du` giving `ln\ u`.]

Now, when `x = 0`, `y = 1`; so we have:

`(-1)/2ln|6-2(1)|=0+K`

`K=(-1)/2ln|4|`

So, on substituting this back into our previous equation and doing some algebra, we obtain:

`(-1)/2ln|6-2y|=x+(-1)/2ln|4|`

`(-1)/2[ln|6-2y|-ln 4]=x`

`ln (6-2y)/4=-2x`

Taking "`e` to both sides":

`(6-2y)/4=e^(-2x)`

`6-2y=4e^(-2x)`

`y=3-2e^(-2x)`

Checking our solution: `(dy)/(dx)=4e^(-2x)`and so

`"LHS"=(dy)/(dx)+2y`

`=4e^(-2x)+2(3-2e^(-2x))`

`=6`

`="RHS"`

Also, when `x = 0`, `y = 3 - 2e^0= 1`.

So the particular solution is given by: `y = 3 - 2e^(-2x)`

Get the Daily Math Tweet!

IntMath on Twitter

Here is the graph of our solution for Example 6:

Solution graph `y=3-2e^(-2x)`, showing the curve passing through `(0, 1)`.

### Example 7

Solve

`x\ dy = y\ ln\ y\ dx`

,

given

`x = 2` when `y = e`.

Answer

Separating variables:

`x\ dy=y\ ln\ y\ dx`

`(dy)/(y\ ln\ y)=(dx)/x`

Integrating: [For the *y* part, let *u* = ln *y*, then *du = dy/y*].

`int(dy)/(y\ ln\ y)=int(dx)/x`

`ln(ln\ y)=ln\ x+K`

Substituting *x* = 2 when *y* = *e* gives:

`ln(ln\ e)=ln\ 2+K`

`ln(1)=ln\ 2+K`

`0=ln\ 2+K`

`K=-ln\ 2`

Substituting this in our general solution:

`ln(ln\ y)=ln\ x-ln\ 2` `=ln\ x/2`

This gives us:

`ln\ y=x/2`

So the particular solution is given by:

`y=e^(x"/"2)`

Here is the graph of our solution for Example 7:

Solution graph `y=e^(x//2)`, showing the curve passes through `(2, e)`.

### Example 8 - RL Circuit Application

In an RL circuit, the differential equation formed using Kirchhoff's law, is

`Ri+L(di)/(dt)=V`

Solve this DE, using separation of variables, given that

R= 10 Ω,L= 3 H andV= 50 volts, andi(0) = 0.

Answer

Substituting *R* = 10, *L* = 3 and *V* = 50 gives:

`10i+3(di)/(dt)=50`

`3(di)/(dt)=50-10i`

First, we separate the variables.

`(di)/((50-10i))=(dt)/3`

Integrate.

`1/10int(di)/(5-i)=1/3intdt`

`-1/10ln(5-i)=t/3+K`

Since *i*(0) = 0,

`-1/10ln(5-0)=0+K`

`K=(-ln\ 5)/10`

So, substituting for *K*:

`-1/10ln(5-i)=t/3-(ln\ 5)/10`

Put log parts together.

`-t/3=1/10ln(5-i)-(ln\ 5)/10`

Multiply both sides by 10

`-(10t)/3=ln(5-i)-ln\ 5`

`-(10t)/3=ln\ (5-i)/5`

`e^(-10t"/"3)=(5-i)/5`

`5e^(-10t"/"3)=5-i`

`i=5-5e^(-10t"/"3)=5(1-e^(-10t"/"3))`

The graph shows that the current builds up and levels out at a maximum value of 5 A.

Solution graph `i=5(1-e^(-10t"/"3))`.

**NOTE:** We could have solved this for `i` another way. Here it is - you may find it easier.

`-1/10ln(5-i)=t/3-(ln\ 5)/10`

`ln(5-i)=-(10t)/3+ln\ 5`

Raising both sides as a power of* e*:

`5-i=e^(-10t"/"3 + ln\ 5)` `=e^(-10t"/"3)e^(ln\ 5)` `=5e^(-10t"/"3)`

`i=5-5e^(-10t"/"3)` `=5(1-e^(-10t"/"3))`

Please support IntMath!

### Example 9 - Skydiver's Terminal Velocity

Image source.

When a skydiver jumps out of a (perfectly good) aeroplane, apart from experiencing sheer terror (and exhilaration), she experiences two forces:

Gravity, acting downwards

Air resistance, acting upwards

Gravity is written *g* and on earth its acceleration has value 9.8 ms^{-2}.

Air resistance depends on the size and shape of the item that is dropping through the air. We use the constant *k* to represent the amount of air resistance. It is called the **coefficient of drag**.

The air resistance is proportional to the **square of the velocity**, so the upwards resistance force due to air resistance can be represented by

F_{air}=kv^{2}.

Now the force on the object due to gravity is:

F_{gravity}=mg, wheregis the acceleration due to gravity.

The total force acting on the body is

`F_("total") = F_("air") - F_("gravity") = kv^2 - mg`

Dividing throughout by *m* gives us a good model for the velocity *v* of an object falling through the air:

`a=k/mv^2-g`

We can write this as

`(dv)/(dt)=k/mv^2-g`

Image source.

**Questions**

a. Find an expression for the velocity of a sky diver at time *t*.

b. Find the velocity after 5 seconds for a sky diver of mass *m* = 80 kg and *k* = 0.2

c. Find the terminal velocity for any object for general values of *g* and *k*.

d. Find the terminal velocity our skydiver with mass *m* = 80 kg and *k* = 0.2.

Answer

#### a. General Formula for Velocity of an Object Falling in Air

We rewrite our differential equation:

`(dv)/(dt)=k/mv^2-g`

in a more convenient form :

`(dv)/(dt)=g(k/(mg)v^2-1)`

Also for convenience, let *c* be defined as:

`c=sqrt((mg)/k`

So:

`1/c^2=k/(mg)`

We can rewrite our differential equation as:

`(dv)/(dt)=g(v^2/c^2-1)`

This is the same as:

`(dv)/(dt)=g((v^2-c^2)/c^2)`

Separating the variables:

`(c^2dv)/(v^2-c^2)=g\ dt`

Integrating both sides:

`c^2int(dv)/(v^2-c^2)=intg dt`

We multiply the fraction by `-1`, thus reversing the order of `(v^2-c^2)`, and also at the front to compensate, so the substitution step coming up later is possible. (Otherwise, we would be trying to find the log of a negative number when finding `K`.)

`-c^2int(dv)/(c^2-v^2)=int g dt`

To perform this integration, we could either:

- Factor the denominator, use partial fractions and then integrate (it needs the logarithm form), or
- Use a table of integrals (integral #13), (easier); or
- Use a computer algebra system, like Scientific Notebook. (easiest)

We integrate and obtain:

`(-c^2)1/(2c)ln((c+v)/(c-v))= g t +K`

"*K*" is the constant of integration.

At `t = 0`, `v = 0` and after substituting, we conclude `K = 0`. Therefore:

`-c/2ln((c+v)/(c-v))=g t`

Now we solve for *v*.

Multiply both sides by `-2` and divide by *c*:

`ln((c+v)/(c-v))=-(2g t)/c`

Take *e* to both sides to remove the logarithm:

`(c+v)/(c-v)=e^(-2g t"/"c)`

Multiply out and solve for *v*:

`c+v=(c-v)e^(-2g t"/"c)`

`c+v=ce^(-2g t"/"c)-ve^(-2g t"/"c)`

`ve^(-2g t"/"c)+v=ce^(-2g t"/"c)-c`

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

This is the velocity of our object at time *t*.

#### b. Velocity for Particular Sky Diver at Specific Time

First, we find *c* for our given situation. This was the expression for *c*:

`c=sqrt((mg)/k)`

We use the given mass and the coefficient of drag for the skydiver.

mass =

m= 80 kgcoefficient of drag =

k= 0.2

So

`c=sqrt((mg)/k)`

`=sqrt((80xx9.8)/0.2)`

`=62.6`

Next, we substitute the given time value (*t* = 5) and *c* = 62.6 to find the required velocity:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

`=62.6(e^(-2(9.8)(5)"/"62.6)-1)/(e^(-2(9.8)(5)"/"62.6)+1)`

`=-40.9580`

The units are ms^{-1}, so the velocity at time *t* = 5 s is approximately 147 km/h (1 ms^{-1} = 3.6 km/h), in the downward direction.

#### c. Terminal velocity

The expression for velocity at time *t* we found earlier:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

As *t* → ∞, the value of the fraction approaches −1, since *e*^{-2gt/c} → 0, giving us the terminal velocity `v=-c`.

So

`c=sqrt((mg)/k)`

is the **terminal velocity** for the falling object (in the downward direction).

**Note: **We could have obtained the above expression without knowing the expression for velocity at time *t*, by simply noting the velocity of the object reaches terminal velocity when the acceleration is 0.

That is, solving the following acceleration expression to find the velocity:

`a=k/mv^2-g=0`

This gives terminal velocity

`v=sqrt((mg)/k)`

#### d. Terminal Velocity for Skydiver Example

We already found the expression for *c* (which is the terminal velocity) in Part (b)

`c=sqrt((mg)/k)`

`=sqrt((80xx9.8)/0.2)`

`=62.6`

So the terminal velocity is approximately 225 km/h, since 1 ms^{-1} = 3.6 km/h.

The graph of the velocity against time shows that it takes around 15 seconds to reach (or "get very close to") the terminal velocity:

Velocity graph `v=225(e^(-2(9.8) t"/"62.6)-1)/(e^(-2(9.8) t"/"62.6)+1)`, showing the point `(5, -147)`.

**Note:**

- The graph shows that the terminal velocity is never actually reached - the skydiver's velocity just gets closer and closer to that velocity.
- The graph includes the point representing the velocity at time
*t*= 5, found earlier. - Actually, the air resistance increases as the air gets more dense nearer the Earth's surface. We have assumed it remains constant for this problem.
- The human sky diver can change
*k*easily by either spreading their arms and legs (which will slow them down), or diving down with arms and legs tightly together (which will increase their speed)

Please support IntMath!

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!