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., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

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:

- $\frac{∂F}{∂x}$ = F
_{x}= M(x, y) - $\frac{∂F}{∂y}$ =F
_{y}= N(x, y)

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 = F_{x}dx + F_{y}dy = 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 F_{x} = M and F_{y} = 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 $\frac{∂F}{∂x}=M(x,y), \frac{∂F}{∂y} = N(x, y).$

- Consider the equation: (2xy -9x
^{2}) + (2y+x^{2}+1)y’ = 0

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: F_{x} = M(x, y) = 2xy -9x^{2} and F_{y} = N(x, y) = 2y+x^{2} +1.

**Integrate M with respect to x**: F_{x}= 2xy -9x^{2}⇒$F = \int (2xy -9x^2)dx = x^2y -3x^3 + g(y)$ where g(y) is the “constant” of integration with respect to x, which may be a function of y.**Differentiate F with respect to y and set it equal to N**: $\frac{∂F}{∂y} = \frac{∂}{∂y}(x^2y -3x^3 + g(y))= 2y + x^2+ 1 ↭ x^2 + g’(y) = 2y + x^2+ 1 ⇒ g’(y) = 2y + 1$**Find g(y) by integrating g’(y)**: g(y) = $\int (2y +1)dy = y^2 + y$. The constant C can be absorbed into the final constant when setting F(x, y) = C, so we can omit it here.**Write the complete expression for F(x, y)**: F(x, y) = $x^2y -3x^3 + g(y) = x^2y -3x^3 + y^2 + y$. The implicit solution is $F(x, y) = x^2y -3x^3 + y^2 + y = C.$ In other words, this equation implicitly defines y as a function of x, representing the general solution to the differential equation.

- sin(x)cos(y)dx + cos(x)sin(y)dy = 0

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: F_{x} = sin(x)cos(y) and F_{y} = cos(x)sin(y).

**Integrate M with respect to x**: F_{x}= sin(x)cos(y) ⇒$F = \int (sin(x)cos(y))dx = -cos(x)cos(y) + g(y)$ where g(y) is a function of y alone (the “constant” of integration with respect to x).**Differentiate F with respect to y and set it equal to N**: $\frac{∂F}{∂y} = \frac{∂}{∂y}( -cos(x)cos(y) + g(y))= cos(x)sin(y) + g’(y) = cos(x)sin(y) ↭ g’(y) = 0$**Find g(y) by integrating g’(y)**: g(y) = $\int 0·dy = C$. Again, the constant C can be absorbed into the final constant when setting F(x, y) = C.**Write the complete expression for F(x, y)**: F(x, y) = $-cos(x)cos(y)$. The implicit solution is $F(x, y) = -cos(x)cos(y) = C.$ We can also write this as F(x, y) = cos(x)cos(y) = D, by absorbing the negative sign into the constant D = −C.

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 μ_{x} = f’(x)g(y) and μ_{y} = 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.

**If $\frac{M_y-N_x}{N}$ is a function of x alone**, then we let μ = μ(x) = $e^{\int \frac{M_y-N_x}{N}dx}$.**If $\frac{N_x-M_y}{M}$ is a function of y alone**, then we let μ = μ(y) = $e^{\int \frac{N_x-M_y}{M}dy}$- If $\frac{N_x-M_y}{xM-yN}$ is a function of xy (meaning it can be written solely in terms of xy), then we let μ = μ(xy) = $e^{\int \frac{N_x-M_y}{xM-yN}(t)dt}$

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 e^{C} is an arbitrary constant, we can absorb it into μ(t). μ(t) = $e^{\int \frac{N_x-M_y}{xM -yN}(t)dt}$.

- Solve $(3xy + y^2) + (x^2+xy)\frac{dy}{dx} = 0$

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 M_{y} = N_{x}: $M_y = 3x + 2y ≠ N_x = 2x + y$. 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 = ±e^{C} 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**

**Integrate M with respect to x**, $\frac{∂F}{∂x} = F_x = 3x^2y + xy^2, F = \int (3x^2y + xy^2)dx = x^3y + \frac{1}{2}x^2y^2 + g(y)$**Differentiate F with respect to y and set it equal to N**: $\frac{∂F}{∂y} = F_y = \frac{∂}{∂y}( x^3y + \frac{1}{2}x^2y^2 + g(y))= x^3 + x^2y + g’(y) = x^3 + x^2y ⇒ g’(y) = 0. $**Find g(y) by integrating g’(y)**: g(y) = $\int 0dy = C$. The constant C can be absorbed into the final constant when setting F(x, y) = C.**Write the complete expression for F(x, y)**: F(x, y) = $x^3y + \frac{1}{2}x^2y^2$. The implicit solution is $x^3y + \frac{1}{2}x^2y^2 = C$.

- $(5xy + 2x +5)dx + 2xdy = 0$

Step 1: **Check for Exactness**

With M = 5xy +2x +5 and N = 2x, compute: M_{y} and N_{x}: $M_y = 5x ≠ N_x = 2$. Since M_{y} ≠ N_{x}, 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 = e^{C}] $k\frac{1}{x}e^{\frac{5}{2}x}$ =[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**

**Integrate N with respect to y**, $\frac{∂F}{∂y} = F_y = 2e^{\frac{5}{2}x}, F = \int (2e^{\frac{5}{2}x})dy = 2ye^{\frac{5}{2}x} + g(x)$**Differentiate F with respect to x and set it equal to M**: $\frac{∂F}{∂x} = F_x = \frac{∂}{∂x}( 2ye^{\frac{5}{2}x} + g(x))= 5ye^{\frac{5}{2}x} + g’(x) = 5ye^{\frac{5}{2}x} +2e^{\frac{5}{2}x} + \frac{5}{x}e^{\frac{5}{2}x} ⇒ g’(x) = +2e^{\frac{5}{2}x} + \frac{5}{x}e^{\frac{5}{2}x}. $**Find g(x) by integrating g’(x)**: g(x) = $\int (2e^{\frac{5}{2}x} + \frac{5}{x}e^{\frac{5}{2}x})dx = \frac{4}{5}e^{\frac{5}{2}x} + 5\int \frac{e^{\frac{5x}{2}}}{x}dx + C$. The constant C can be absorbed into the final constant when setting F(x, y) = C, and the integral $5\int \frac{e^{\frac{5x}{2}}}{x}dx$ does not have an elementary antiderivative, so we leave it in integral form:**Write the complete expression for F(x, y)**: F(x, y) = $2ye^{\frac{5}{2}x} + \frac{4}{5}e^{\frac{5}{2}x} + 5\int \frac{e^{\frac{5x}{2}}}{x}dx$. The implicit solution is $F(x, y) = 2ye^{\frac{5}{2}x} + \frac{4}{5}e^{\frac{5}{2}x} + 5\int \frac{e^{\frac{5x}{2}}}{x}dx = C$.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: $\int \frac{e^{\frac{5x}{2}}}{x}dx = Ei(\frac{5}{2}x)+C’$.

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