All those who seem stupid, they are, and also so are half of those who do not, Quevedo
I will be dust, but dust in love, Quevedo
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
An exact differential equation is a type of first-order differential equation that can be written in the form: M(x, y) + N(x, y)y' = 0 or equivalently $M(x, y) + N(x, y)\frac{dy}{dx} = 0$ or M(x, y)dx + N(x, y)dy = 0 where M(x, y) and N(x, y) are functions of x and y, and y’ = $\frac{dy}{dx}$ represents the derivative of y with respect to x.
This equation is called exact if there exists a function F(x,y), called the potential function, such that:
In other words, F(x, y) is a function whose total differential matches the expression in the differential equation. Specifically, we have: $dF = \frac{∂F}{∂x}dx + \frac{∂F}{∂y}dy =[\text{Notation}] F_xdx + F_ydy = M(x, y)dx + N(x, y)dy$
F(x, y) is a potential function whose total differential, dF = Fxdx + Fydy = Mdx + Ndy, equals zero when set to the differential equation Mdx + Ndy = 0. The solution to the differential equation is then implicitly given by F(x, y) = C, where C is constant.
To confirm that a differential equation Mdx + Ndy = 0 is exact, we check the condition: $\frac{∂M}{∂y} = \frac{∂N}{∂x}$. If this condition is satisfied, then the equation is indeed exact, and there exists a potential function F(x, y) such that Fx = M and Fy = N.
Definition. We say a differential equation is exact if it can be written as M(x, y) + N(x, y)y' = 0, and if it satisfies $\frac{∂N}{∂x}=\frac{∂M}{∂y}$. Then, a solution y(x) is then implicitly described by F(x, y) = C where $\frac{∂F}{∂x}=M(x,y), \frac{∂F}{∂y} = N(x, y).$
We rewrite it as: $(2xy-9x^2)dx + (2y+x^2+1) = 0$ where M(x, y) = $2xy-9x^2$ and N(x, y) = $2y+x^2+1$.
To verify if the differential equation is exact, we need to check whether: $\frac{∂}{∂x}(2y+x^2+1) = 2x = \frac{∂}{∂y}(2xy-9x^2)$. Since $\frac{∂N}{∂x} = 2x = \frac{∂M}{∂y}$, the equation is exact.
We aim to find a function F(x, y) such that: Fx = M(x, y) = 2xy -9x2 and Fy = N(x, y) = 2y+x2 +1.
Here, M(x, y) = sin(x)cos(y) and N(x,y) = cos(x)sin(y).
Check for Exactness: $\frac{∂}{∂x}(cos(x)sin(y)) = -sin(x)sin(y) = \frac{∂}{∂y}(sin(x)cos(y)$. This is a exact differential equation.
We seek a function F(x, y) such that: Fx = sin(x)cos(y) and Fy = cos(x)sin(y).
When we encounter a non-exact differential equation, it is sometimes possible to transform it into an exact one by finding an appropriate integrating factor, denoted as μ(x,y). Multiplying the entire equation by μ mat make it exact, allowing us to apply methods for solving exact equations.
Given a differential equation of the form: M(x, y)dx + N(x, y)dy=0, we can attempt to find a suitable integrating factor μ such that: μMdx + μNdy = 0 is exact.
For the equation to be exact after multiplication, it must satisfy the following condition: $\frac{∂}{∂y}(μM) = \frac{∂}{∂x}(μN) ↭[\text{which expands as}] μ_yM + μM_y = μ_xN + μN_x ↭[\text{Simplifying, we get}] μ_yM -μ_xN = μ(N_x -M_y)$
Assuming μ(x, y) = f(x)·g(y), we then have μx = f’(x)g(y) and μy = f(x)g’(y), and the condition $μ_yM -μ_xN = μ(N_x -M_y)$ simplifies to: $f(x)g’(y)M -f’(x)g(y)N = f(x)g(x)(N_x -M_y)$ ⇒[Let’s divide both terms by 1⁄μ] $\frac{g’(y)}{g(y)}M -\frac{f’(x)}{f(x)}N = N_x -M_y$
If μ = μ(x), i.e., f(x) = μ(x), g(y) = 1 ⇒ g’(y) = 0.
Using these new restrictions, we get: $-\frac{f’(x)}{f(x)}N = N_x -M_y ↭ \frac{μ’}{μ} = \frac{M_y-N_x}{N}$. By assumption μ = μ(x), so the left hand side of the equation is a function of x, then the right hand side of the equation, $\frac{M_y-N_x}{N}$, is also a function of x.
Let’s integrate: $\int \frac{μ’}{μ}dx = \int \frac{M_y-N_x}{N}dx ↭ ln(μ) = \int \frac{M_y-N_x}{N}dx ↭ μ = e^{\int \frac{M_y-N_x}{N}dx}$
The differential equation M + Ny' = 0 ↭ Mdx + Ndy = 0 can be transformed to an exact differential equation μM + μNy' = 0.
Case 3. μ = μ(xy)
$\frac{∂}{∂x}μ = μ’(xy)\frac{∂}{∂x}(xy) = μ’y, \frac{∂}{∂y}μ = μ’(xy)\frac{∂}{∂y}(xy) = μ’x$
Mdx + Ndy = 0 is exact ↭ $\frac{∂}{∂x}(μN) = \frac{∂}{∂y}(μM) ↭ μ’yN + μN_x = μ’xM + μM_y ↭ μ(N_x-M_y) = μ’(xM -yN) ↭ \frac{μ’}{μ} = \frac{N_x-M_y}{xM -yN}⇒ \int \frac{μ’}{μ}(t)dt = \int \frac{N_x-M_y}{xM -yN}(t)dt ⇒ ln(μ(t)) = \int \frac{N_x-M_y}{xM -yN}(t)dt + C ⇒ μ(t) = e^Ce^{\int \frac{N_x-M_y}{xM -yN}(t)dt}$. Since eC is an arbitrary constant, we can absorb it into μ(t). μ(t) = $e^{\int \frac{N_x-M_y}{xM -yN}(t)dt}$.
Rewrite this equation as: $(3xy + y^2)dx + (x^2+xy)dy = 0$ where M = $3xy + y^2$ and N = $x^2+xy$
Step 1: Check for Exactness. Check if My = Nx: $M_y = 3x + 2y ≠ N_x = 2x + y$. We don’t have an exact equation 🤔, but it’s still possible that we can transform it to an exact differential equation.
Step 3: Find an Integrating Factor: Compute $\frac{M_y-N_x}{N} = \frac{3x + 2y -(2x + y)}{x^2+xy} = \frac{x + y}{x(x +y)} = \frac{1}{x}$ which depends only on x.
This suggests using an integrating factor $μ = μ(x) = e^{\int \frac{M_y-N_x}{N}dx} = e^{\int \frac{1}{x}dx} = e^{ln|x| + C} = e^C·|x|$ =[Let k = ±eC is a constant] kx =[Since integrating factors are determined up to a multiplicative constant, we can set k = 1 without loss of generality] x.
Step 3: Multiply by the Integrating Factor μ(x) and simplify.
$μM(x, y)dx + μN(x, y)dy = 0 ↭ x(3xy + y^2)dx + x(x^2+xy)dy = 0 ↭ (3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0$
Step 4: Verify Exactness of the Transformed Equation: $\frac{∂}{∂y}(3x^2y + xy^2) = 3x^2 + 2yx = \frac{∂}{∂x}(x^3 + x^2y)$
Step 5. Solve the Exact Differential Equation
Step 1: Check for Exactness
With M = 5xy +2x +5 and N = 2x, compute: My and Nx: $M_y = 5x ≠ N_x = 2$. Since My ≠ Nx, the equation is not exact, but it’s still possible that we can transform it to an exact differential equation.
Step 2. Find an Integrating Factor:
Compute $\frac{M_y-N_x}{N} = \frac{5x -2}{2x} = \frac{5}{2} - \frac{1}{x}$ which is a function of x. This suggests $μ = μ(x) = e^{\int \frac{M_y-N_x}{N}dx} = e^{\int \frac{5}{2} - \frac{1}{x}dx} = e^{\frac{5}{2}x -ln(x) + C} = e^{\frac{5}{2}x +ln(\frac{1}{x}) + C} = e^{ln(\frac{1}{x})}e^{\frac{5}{2}x}e^C$ =[let k = eC] $k\frac{1}{x}e^{\frac{5}{2}x}$ =[Since integrating factors are determined up to a multiplicative constant, we can set k = 1 without loss of generality] $\frac{1}{x}e^{\frac{5}{2}x}$.
Step 3. Multiply our differential equation by the integrating factor μ(x) and simplify.
$(5ye^{\frac{5}{2}x} +2e^{\frac{5}{2}x} + \frac{5}{x}e^{\frac{5}{2}x})dx + 2e^{\frac{5}{2}x}dy = 0$
Step 4. Solve the Exact Differential Equation
This equation implicitly defines y as a function of x. The integral $\int \frac{e^{\frac{5x}{2}}}{x}dx$ does not have an elementary antiderivative and must remain as is or you can express it in terms of the Exponential Integral function: $\int \frac{e^{\frac{5x}{2}}}{x}dx = Ei(\frac{5}{2}x)+C’$.