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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