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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Resolution of Exact Differential Equations

Let's consider a first order differential equation in the form M(x,y) + N(x,y)dy/dx = 0 or M(x,y)dx + N(x,y)dy = 0. The equation is exact if Mₓ(x,y) = Nᵧ(x,y). For simplification we write Mₓ = Nᵧ. Mₓ is the partial derivative with respect to x. Nᵧ is the partial derivative with respect to y.

In this case there will be a function ψ(x,y) such that ψₓ = M and ψᵧ = N where the equation ψ(x,y) = c will implicitly define a function that will satisfy the original differential equation.

Example: Solve the differential equation 2xy - 9x² + (2y + x² + 1)dy/dx = 0 using the initial condition y(0) = -3

Let's see if the differential equation is exact. 

Let's write M = 2xy - 9x² and N = 2y + x² + 1

Let's take the partial derivatives of these equations:

Mₓ = 2x and Nᵧ = 2x

The differential equation is exact since Mₓ = Nᵧ

There will be a function ψ(x,y) such that ψₓ = M and ψᵧ = N where the equation ψ(x,y) = c will implicitly define a function that will satisfy the original differential equation.

Let's work with ψₓ = M to find ψ (pronounce psi)

Let's substitute M:

 ψₓ = 2xy - 9x²

dψ/dx = 2xy - 9x²

dψ = (2xy - 9x²)dx

∫dψ = ∫(2xy - 9x²)dx

Since we integrate with respect to x, y is constant. That's the reason why we write 2y before the integral sign.

ψ = 2y∫xdx - 9∫x²dx

ψ = (2y)(x²/2) - 9x³/3 + h(y).  Here the constant of integration is a function of y since we integrate with respect to y.

ψ =x²y - 3x³ + h(y)

Let's find h(y) using  ψᵧ = N

Let's find ψᵧ first by differentiating the function ψ with respect to y. Here x² is a constant

ψᵧ = x² + h'(y)

Substitute ψᵧ and N in the expression  ψᵧ = N

x² + h'(y) = 2y + x² + 1

h'(y) = 2y + 1

d[h(y)]/dy = 2y + 1

d[h(y)] = (2y + 1)dy

Let's integrate both sides:

∫ d[h(y)] = ∫ (2y + 1)dy

                   = 2y²/2 + y + k

h(y) =  y² + y + k

Let's substitute h(y) in ψ 

ψ = x²y - 3x³ + y² + y + k

Since ψ = c we have:

c =  x²y - 3x³ + y² + y + k

x²y - 3x³ + y² + y = c-k

Let's just use c and drop k:

x²y - 3x³ + y² + y = c

x²y - 3x³ + y² + y - c = 0

Let's arrange the equation to have a quadratic equation in y:

 y²  +  x²y + y - 3x³ -c = 0

 y²  + ( x² + 1)y - 3x³ - c = 0

Let's find c using the initial condition y(0) = -3

(-3)² + [(0)² + 1](-3) -3[(0)3] - c = 0

Solving this expression we find c = 6

Let's substitute c in the equation:

y²  + ( x² + 1)y - 3x³ - 6 = 0

This equation represents the implicit solution of the differential equation. Let's solve to find the explicit solution

Using the formulas to solve a quadratic equation we have:

y = - (x² + 1)- ⎷ (x² + 1)²-4( - 3x³ - 6 )/2. The whole expresion is under the radical and the expresion - (x² + 1)- ⎷ (x² + 1)²-4( - 3x³ - 6 ) has 2 for denominator. The symbol software I am using doesn't have the horizontal bar for radical and the vertical bar for fraction.

y =  - (x² + 1) + ⎷ (x² + 1)²-4( - 3x³ - 6 )/2

Let's develop the first value of y:

y = - (x² + 1) - ⎷x⁴ + 12x³ + 2x² + 25/2

The second value of y is:

y =  - (x² + 1) + ⎷x⁴ + 12x³ + 2x² + 25/2

Let's see which expression of y satisfies the initial condition:

Substitute y by -3 and x by 0 in the first value of y:

-3 = -[ (0)² + 1] -⎷(0)⁴ + 12(0)³ + 2(0)² + 25/2

-3 = -1 - ⎷25/2 = -1 - 5/2 = -6/2 = -3

The first value of y satisfies the initial condition

Let's see if the second value of y satisfies the initial condition:

Let's substitute x and y in the second value of y:

3 = -[ (0)² + 1] +⎷(0)⁴ + 12(0)³ + 2(0)² + 25/2

-3 = -1 + 5/2 = 4/2 = 2

It doesn't satisfy the initial condition. Therefore only the first value of y is the solution of the differential equation. The explicit solution is : y = - (x² + 1) - ⎷x⁴ + 12x³ + 2x² + 25/2.

The solution exists if x⁴ + 12x³ + 2x² + 25>0. To study the sign of this expression we will have to factorize it or we can graph it using a programmable calculator. The graph will allow to study the sign. Thus the domain of validity of the solution of the differential equation can be set. But here they didn't ask to find the domain of validity.

Practice. Solve the same differential equation without using the initial condition.

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Resolution of a differential equation of first order by the method of separable variables using a given condition

 The method is the same as in resolving a differential equation without a given condition. The only difference here is to find the value of k for the given condition.

Example: Solve the following differential equation and determine the interval of validity for the given solution: dy/dx = 6y²x     y(1) = 1/25

Solution

Let's separate the variables by multiplying both sides by 1/y²:

(1/y²) (dy/dx) = (1/y²) (6y²x)

                      =  6x

Multiply both sides by dx:

1/y² dy = 6xdx

Integrate both sides:

∫ 1/y² dy = ∫6xdx

y^-1/-1 = 6x²/2 + K

-1/y = 3x² + K

-1 = y(3x² + K)

y = -1 /3x² + K

Substitute y by 1/25 and x by 1:

1/25 = -1/[3(1) +K]

1/25 = -1/3 + K

3 + K = -25

k = -25-3 k = -28

Substituting k in y we have:

y = -1/3x²-28

y exists if 3x²-28 is different of zero.

Let's study the sign of  3x²-28. In order to do that we need to equal this expression to 0 and solve the equation.

3x²-28 = 0

By solving this equation we find x = -2⎷7/3 and x = 2⎷7/3

x -∞                        -2⎷7/3            2⎷7/3            +∞

y                 +            0         -           0         +

The intervals in which 3x²-28 is different of zero are:

 -∞ < x < -2⎷7/3   -2⎷7/3 < x < 2⎷7/3  and  2⎷7/3  < x < +∞

These determine the intervals of validity of the solution of the differential equation.

Resolution of differential equations of first order by the method of separable variables

 A non-linear first order differential equation is separable if it can be written in this form:

N(y)dy/dx = M(x)

In order for such a differential equation to be separable, all the y's must be in the first side and all the x's must be in the second side. The independent variable can be in the second side and the dependent variable on the first side. As long as the variables are separated. it doesn't matter. The variable y and x should not be together. They should be separated with one variable in one side and the other in the other side.

Let's integrate both sides of the equation:

∫N(y)(dy/dx).dx = ∫M(x)dx

Let's do a change of variable: u = y(x) . Then we have:

du/dx = y'(x)  

          = dy/dx

du = dy/dx.dx

Substituting y and (dy/dx)dx we have:

∫N(u)du = ∫M(x)dx

In order to solve this equation we have to solve the second integral and substitute for u in the first one.

Let's say that instead of the variable u, we have the variable y the above equation can be written as: ∫N(y)dy = ∫M(x)dx

The solution of this equation can lead to the solution of the differential equation. After calculating the two integrals we arrive to an implicit solution that can lead to an explicit one. An implicit solution is a solution where the value of isn't yet found.

Example: Solve the differential equation dy/dx = (x² - 4) (3y+2)

Let's separate the variables by having y and its derivative in the first side and the x in the second side. We divide both sides by 3y + 2 in order to have x in the second side and y in the first side

(dy/dx)(1/3y+2) = x² - 4

Let's multiply  both sides by dx and simplify by dx in the first side

dx(dy/dx) (1/3y+2) = dx(x² - 4)

dy/3y + 2 = dx(x² - 4)

Let's integrate both sides:

∫dy/ (3y+2) = ∫(x² - 4)dx

The derivative of the numerator being near of that of the numerator, we can do a change of variable by writing 3y + 2 = u. By derivating this expression we get:

du/dy = 3  dy = 1/3du

The above equality becomes by substituting dy and du

∫(1/3du)/u  = 1/3x³- 4x + C

∫(1/3du) (1/u) = 1/3x³- 4x + C

1/3∫ du/u =  1/3x³- 4x + C 

1/3lnu + K = 1/3x³- 4x + C  with u>0 because we can't take the log of negative numbers

Let's substitute u and reduce to the same denominator i:

1/3ln(3y + 2) + K = 1/3x³ - 12x/3 + 3C/3

Let's simplify by 3:

ln(3y + 2) + 3k = x³ - 12x + 3C

ln(3y + 2) = x³ - 12x + 3C-3K

We substitute 3C-3K by another variable

ln(3y + 2) = x³ - 12x + K₁

By definition of logarithm, if lnx = y, then logₑx = y. Therefore e^y = x

The above equation becomes e^(x³ - 12x + K₁ ) = 3y + 2

y = 1/3[ e^(x³ - 12x +  K₁) - 2]

Practice

Solve dy/dx = 2xy +3y -4x-6