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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