Bernoulli Differential Equations

  Let's consider a differential equation in the form y' + p(x)y = q(x)yⁿ where p(x) and q(x) are continuous and n a real number. Such an equation is called Bernoulli equation. If n = 0 or n = 1, the equation becomes easy to solve as an ordinary differential equation.

We are going to use the substitution method to solve this equation. It appears that the term yⁿ is a problem. Let's get rid of it by multiplying both sides by its inverse.

y⁻ⁿy' +y⁻ⁿp(x)y =  q(x)yⁿy⁻ⁿ

y⁻ⁿy' + p(x)y¹⁻ⁿ = q(x)

Now let's do the substitution that will lead us to an equation easier to solve. Let's write v = y¹⁻ⁿ 

Let's differentiate both sides. v is a function of y and y is a function of x. We use the derivative formula of uⁿ = nuⁿ⁻¹u' to find the derivative of  y¹⁻ⁿ 

v' = (1-n) y¹⁻ⁿ⁻¹y'

y' = v'/(1-n) y¹⁻ⁿ⁻¹ = v'/(1-n)y⁻ⁿ

Now if we substitute y' and v in the equation we will have an equation in v easier to solve. Let's do that.

y⁻ⁿv'/(1-n)y⁻ⁿ + p(x)v = q(x)

Let's simplify the first expression by y⁻ⁿ:

v'/(1-n) + p(x)v = q(x)

This is a linear differential equation in v that can be solved by finding v. Once we find v we can substitute it and find y. Let's do a numerical equation.

Example

Let's solve the equation: y' + (4/x)y = x³y² 

Let's multiply both sides by y⁻² to get rid of y² :  y⁻²y' + (4/x)yy⁻² = x³y² y⁻² (1)

y⁻²y' + (4/x)y⁻¹ = x³ (2)

Let's write v = y⁻¹

Let's derive both sides:

v' = (-1)y⁻¹⁻¹y'

v' = -y⁻²y'

Let's find y': 

y' = v'/-y⁻²

y' = - v'y²

Let's substitute v and y' in the equation (2):

(y⁻²)( - v'y²) + (4/x)v = x³

- v' +  (4/x)v = x³

v' -(4/x)v = -x³ (3)

To solve this linear differential equation in v we have to find μ:

μ = e^∫p(x)dx

p(x) is the coefficient of v: p(x) = -4/x

Let's calculate ∫p(x)dx

∫p(x)dx = ∫-4/xdx = -4lnx = lnx⁻⁴

Let's substitute ∫p(x)dx  in the formula of μ:

μ = e^ lnx⁻⁴ = x⁻⁴

Let's multiply both sides of equation (3) by x⁻⁴:

v'x⁻⁴ - x⁻⁴.(4/x)v = (x⁻⁴)( -x³)

v'x⁻⁴ - 4x⁻⁵v = -x⁻¹

The left side is the derivative of (x⁻⁴v):

(x⁻⁴v)' = -x⁻ ¹

Let's use the notation d/dx:

d(x⁻⁴v)/dx = -x⁻¹

d(x⁻⁴v) = -x⁻¹dx

Let's integrate both sides:

∫d(x⁻⁴v) = ∫-x⁻¹dx

x⁻⁴v = -∫1/xdx = -lnx + k

v = (-lnx + k)/x⁻⁴

v = x⁴(-lnx + k)

Let's substitute v in v = y⁻¹:

 x⁴(-lnx + k) = y⁻¹:

y = 1/x⁴(-lnx + k)

This value exists only if x is different of 0.

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Derivative of logarithmic functions

In this post I'll show some techniques to remember the formulas for logarithmic  functions. I'll do some examples and leave some exercises to practice.

Derivative of logarithmic functions 

Derivative of log_bx

d/dx (log_bx) = 1/xlnb
To remember this formula let's apply the following technique:
1) Multiply the number of which we calculate the logarithm by the natural logarithm of the base. The number here is x and the base is b. Therefore we have xlnb
2) Take the inverse of this product. The inverse of the product is 1/xlnx

Derivative of lnx

d/dx(lnx) = 1/x
The derivative of the logarithm of any number is equal to the inverse of this number.

Derivative of log_bu

Since log_bu is a composite function its derivative is given by d/dx(log_bu) = d/du(log_bu).du/dx

d/dx(log_bu) = 1/ulnnb.du/dx

Rule: The derivative of the logarithm of a composite function is equal to its derivative with respect to the new variable (u) multiplied by the derivative of the new variable (u) with respect to x.

Derivative of lnu 

Since u is a composite function we have d/dx(lnu) = d/du(lnu).du/dx
                                                                              = i/u.du/dx

Rule: The derivative of the natural logarithm of a composite function u is equal to the inverse of the function multiplied by its derivative with respect to x

Example 1, Calculate the derivative of y = x³log₅2x

The derivative of y is y" = (x³log₅2x)'

Let's apply the product rule:
Y' = (x³)'(log₅2x) + x³(log₅2x)'
The derivative of x³ is obvious. Let's calculate the derivative  of log₅2x
Let's write u = 2x we have (log₅u)' = d/du(log₅u),du/dx
                                                     = i/uln5.u'
                                                    = 1/2x.ln5.(2x)'
                                                   = 1/2x.ln5.2
                                                   = 1/xln5
let's go back to the derivative of y we have:
y' = 3x²log₅2x + x³.1/xlnx
   = 3x²log₅2x+x²/lnx

Example 2.  Calculate the derivative of y = ln(2x²-4x+3)

Let's write u = 2x²-4x+3
We have y = lnu
Then dy/dx = d/dx(lnu)
Since lnu is a composite function then dy/dx = d/du(lnu).du/dx
                                                                      = 1/u(4x-4)
Substitute u: dy/dx = (1/2x³-4x+3).(4x-4)
dy/dx = 4x-4/2x²-4x+3
           = 4(x-1)/2x³-4x+3

Practice

Calculate the derivative of the following functions;
1.log₅(2x+5)
2. 5/log(x+4)
3. ln(sinx)

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