Resolution of Differential equations having the form y' = G(ax + by)

This is another form of differential equation that can be solved using the substitution method. Remember that the substitution method was used in the case of homogeneous differential equations. Homogeneous equations have the form y' = F(y/x).

In order to solve the differential equation  y' = G(ax + by), we use the substitution v = ax + by. Then we have y' = G(v)

Let's derive the equation   v = ax + by

v' = a + by' 

by' = v' - a

y' =  (v' - a)/b

Let's equal both expressions of y'

 (v' - a)/b = G(v)

v' - a = bG(v)

v' = a + bG(v)

dv/dx = a + bG(v)

1/dx = a + bG(v)/dv

dx = [1/a + bG(v)]dv

Let's integrate both sides:

∫dx = 1/a + bG(v∫dv

The resolution of this equation allows to find v. Once we have v we can substitute it in v = ax + by that allows to find y.

Example. Solve the differential equation: y' - (4x - y + 1)² = 0

Let's write v = 4x - y

Then the differential equation becomes: y' - (v + 1)² = 0 (1)

Let's derive v in order to obtain y'. Once we find y' we will have a differential equation in v. We will have to solve this equation for v. Once we find v we can substitute to find y.

v' = 4 - y' then y' = 4 - v'.

Let's substitute y' in (1):

4 - v' - (v + 1)² = 0

4 - (v + 1)² = v'

dv/dx = 4 - (v + 1)²

1/dx = 4 - (v + 1)²/dv

dx = dv/4 - (v + 1)²

Let's integrate both sides:

∫dx = ∫dv/4 - (v + 1)² (2)

Let's calculate the second side. In order to do this, let's transform 1/4 - (v + 1)²:

1/4 - (v + 1)² = 1/-v² -2v + 3

                      = -1/v² + 2v -3

                      = - 1/(v + 3) (v - 1)

   Then the equation (2) becomes:

∫dx = - ∫1/(v + 3) (v - 1) dv (3)

Let's develop 1/(v + 3) (v - 1) in partial fractions:

1/(v + 3) (v - 1) = A/v + 3 + B/v-1

By developing this expression we find A = -1/4 and B = 1/4

By substituting we find :

1/(v + 3) (v - 1) = -1/4( v + 3)  + 1/4(v - 1)

The equation (3) becomes:

 ∫dx  = ∫[ 1/4( v + 3)  - 1/4(v - 1)]dv

 ∫dx = 1/4∫(1/v + 3 -1/v - 1)

x + c = 1/4 ln( v + 3/v - 1) (we don't consider a constant of integration for this integral)

4x + 4c = ln( v + 3/v - 1)

v + 3/v - 1 = e^4x + 4c

                 = e^4x.e^4c

                = (e^4x.)(c) ( we consider 4c to be equivalent to another constant c which we can call also c')

v + 3 = ce^4xv - ce^4x

By doing all the calculations, we find v = c^4x + 3/ce^4x - 1

By substituting v in the expression v = 4x - y we find y = 4x - 3 + ce^4x/-1 + ce^4x

Practice. Solve the differential equation y' = e^9y-x

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Homogeneous Differential Equations

 An homogeneous differential equation is one that can be written as y' = F(y/x). The equation is not written in this form most of the times. You have to transform it in order to write it under this form. This type of equation is solved by substitution. That's another type of equation that's solved by substitution. Remember that in the case of a Bernoulli differential equation we use the substitution v = y¹⁻ⁿ. In the case of an homogenous differential equation we are going to see what type of substitution we can use.

Let's bring this homogeneous differential equation to us in order to solve it. Here it comes as y' = F(y/x). It seems natural that we can substitute y/x by something. Let's call it v. We can write v =  y/x. Then we have: y = vx.

Let's derive both sides of the equation:

y' = v'x + v

Remember also that the differential equation is y' = F(y/x) or y' = F(v)

 Equaling both expressions of y' we have:

v'x + v = F(v)

We have a differential equation with a new dependent variable v. This equation can be solved by the method of separation of variables. Let's continue;

v'x = F(v) - v

(dv/dx)x = F(v) - v

x/dx = F(v) - v/dv

dx/x  = dv/F(v) - v

Now that the variables are separated we can integrate both sides of the equation in order to find v. Once we find v we can use the substitution to find y.

Example. Solve the differential equation xyy' + 4x² + y² = 0

Let's bring the expression y/x in the equation by dividing by x. Let's divide both sides by x²:

 xyy'/x² + 4x²/x² +  y²/x²  = 0

yy'/x + 4 + (y/x)² = 0

y/xy' + 4 + (y/x)² = 0

Let's substitute y/x by v:

vy' + 4 + v² = 0  (1)

Let's find y' in order to have an equation with the new variable v:

y/x = v 

y = vx

y' = v'x + v

Let's substitute y' in equation (1)

v(v'x + v) + 4 + v² = 0

vv'x + v² + 4 +  v² = 0

vv'x + 2v² + 4 = 0

vv'x  = -2v² - 4 

Let's divide both sides by v in order to separate the variables:

v'x  = -2v² - 4/v

(dv/dx)x = -2v² - 4/v

x/dx = -2v² - 4/vdv

dx/x = -vdv/ 2v² + 4

Now that the variable are separated let's integrate both sides in order to find v;

∫dx/x = ∫(-v/ 2v² + 4)dv = -∫(v/ 2v² + 4)dv (2)

Let's solve the integral in the second side by doing the substitution u =  2v² +4

du = 4vdv dv = du/4v

The integral at the second side becomes:

∫(v/ 2v² +4)dv = ∫(v/u)(du/4v) = ∫du/4u = 1/4∫du/u = 1/4lnu = 1/4ln (2v² +4) + k

Let's substitute the value of the integral we just calculated in (2)

∫dx/x = -1/4ln (2v² + 4) + k

lnx + K' = -1/4ln (2v² +4) + k

1/4ln (2v² + 4) = -lnx - k' + k

Let's call c the expression-k'+k 

1/4ln (2v² + 4) = -lnx + c

1/4ln (2v² + 4) = -lnx + c

ln(2v² + 4)^1/4  =  -lnx + c

(2v² + 4)^1/4 = e^(-lnx + c)

                     = (e^-lnx)(e^c)

                     = (e^lnx^-1)(d) We call e^c d

  e^lnx^-1 being equal to x^-1, we have:

(2v² + 4)^1/4 = (x^-1)(d) 

Let's raise both sides to the power of 4:

2v² + 4 = x⁻⁴d⁴

Let's call d⁴ g:

2v² + 4 = x⁻⁴g

            = g/x⁴

v²  = g-4x⁴/2x⁴

v = ∓(⎷2/2) (⺁g-4x⁴)/x²

By substituting v in the expression v = y/x, we obtain:

y = ∓⎷2/2⺁g-4x⁴/x

Practice. Solve the differential equation for the initial condition y(2) = -7 and x>0

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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