Solving First-Order Linear ODEs. Integrating Factors

Arithmetic is where numbers fly like pigeons in and out of your head, Carl Sandburg

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It answers two fundamental questions: Is there a solution? and Is it the only one? It states that if:

• The function $f(x, y)$ (the right-hand side of the ODE when written as $y' = f(x, y)$) is continuous in some neighborhood around a specific point $(x_0, y_0)$ and
• Its partial derivative with respect to y, denoted as $\frac{\partial f}{\partial y}$, is also continuous near $(x_0, y_0)$.

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it perfectly satisfies the initial condition $y(x_0) = y_0$.

This theorem is a mathematician’s safety net: it ensures that under these well-behaved conditions, a solution definitively exists and is absolutely unique near $x = x_0$. The physical system modeled by the ODE will only follow one distinct path.

Standard Form

To solve these equations systematically, the general first-order linear ODE is usually rearranged by dividing the entire equation by $a(x)$ (assuming $a(x) \neq 0$), yielding the standard linear form: $\boxed{y' + p(x)y = q(x)}$ where:

• $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$
• a(x) ≠ 0 for all x in the interval of interest.

Expressing the equation in this standard form is the critical first step before applying advanced solution techniques, such as the method of integrating factors.

Solving First-Order Linear ODEs Using Integrating Factors

The method of integrating factors is a common and brilliant algebraic technique for solving first-order linear ODEs in standard form.

Given the equation in standard form: $y' + p(x)y = q(x)$, we define an integrating factor, denoted as $\mu(x)$, calculated as: $μ(x)=e^{\int p(x)dx}$

Why do we do this? Multiplying both sides of the differential equation by $\mu(x)$ “magically” transforms the entire left-hand side into the exact derivative of the product $(\mu(x)y)$, via the Reverse Product Rule. This allows us to simply integrate both sides to find $y(x)$.

Solving First-Order Linear Ordinary Differential Equations (ODEs)

In this section, we will solve several first-order linear ODEs using the method of integrating factors.

• Solving xy’ - y = x³

1. Rewrite the differential equation in standard form (since we are considering x > 0 to avoid division by zero): $y' -\frac{1}{x}y = x^2$, where $p(x) = \frac{-1}{x}, q(x) = x^2.$
2. Calculate the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{-1}{x}dx}$ =[The absolute value ln|x| ensures the logarithm is defined for x < 0 as well, but for simplicity, we assume x > 0] $e^{-ln(x)} = e^{ln(x^{-1})} = x^{-1} = \frac{1}{x}$.

   By the power rule of logarithms log(a^b) = b*log(a). In our example, $e^{-ln(x)} = e^{ln(x^{-1})}$.

3. Multiplying both sides of the differential equation by the integrating factor μ(x): $\frac{1}{x}y'-\frac{1}{x^2}y = x$
4. Recognize the left-hand side as a product derivative μ(x)y, $\frac{d}{dx}(\frac{1}{x}y) = \frac{1}{x}y'-\frac{1}{x^2}y$.
5. Integrate both sides of the equation with respect to x: $\frac{1}{x}y = \int xdx = \frac{x^2}{2} + C$ where C is the constant of integration.
6. Solve for y, $\boxed{y(x) = \frac{x^3}{2}+Cx}$ where C is an arbitrary constant determined by initial conditions (if any are provided).

• Solve $x^2y' + xy = 3x$

Step 1: Rewrite in standard form

This equation is not immediately in the standard linear form, so we will manipulate it to fit that form and then apply the method of integrating factors.

To bring the equation into the standard linear form y′ +P(x)y=Q(x), we divide both sides by x² (with $x \neq 0$): $\frac{1}{x^2}(x^2y' + xy) = \frac{1}{x^2}(3x) ↭ y' + \frac{1}{x}y = \frac{3}{x}$

Step 2: Calculate the integrating factor $\mu(x)$

Compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int \frac{1}{x}dx} = e^{ln|x|} = |x|$. Since x is in the denominator and we need x ≠ 0, we’ll consider x > 0 for simplicity, so μ(x) = x.

Step 3: Multiply and Compress

We write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $xy' + y = 3 ↭ \frac{d}{dx}(x·y) = 3$.

Step 4: Integrate

Next, we can solve the differential equation integrating with respect to x, $xy = \int 3dx = 3x + C$

Step 5: Solve for y

$\boxed{y = 3 + \frac{C}{x}}$ where C is the constant of integration determined by initial conditions.

• Solve (Initial Value Problem) (1+cos(x))y’ -sin(x)y = 2x with y(0) = 1

1. Rewrite the differential equation in standard form: $y' -\frac{sin(x)}{1+cos(x)}y = \frac{2x}{1+cos(x)}$ where $p(x) = -\frac{sin(x)}{1+cos(x)}, q(x) = \frac{2x}{1+cos(x)}$
2. Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int -\frac{sin(x)}{1+cos(x)}dx} = e^{ln(1+cos(x))} = 1 + cos(x)$.

   $\int -\frac{sin(x)}{1+cos(x)}dx$ =[Let u = 1+cos(x), then du = −sin(x)dx] $\int \frac{du}{u} = ln|u| = ln|1+cos(x)|=$[Since 1 + cos(x) ≥ 0 for all real x, we can drop the absolute value:] ln(1 + cos(x))

3. Multiplying both sides of the differential equation by the integrating factor μ(x): $(1+cos(x))y' -sin(x)y = 2x.$
4. By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $( (1+cos(x))\cdot y)' = 2x$
5. Integrate both sides of the equation with respect to x: $(1+cos(x))\cdot y = \int 2xdx = x^2 + C$ where C is the constant of integration.
6. Solve for y, $y = \frac{x^2+C}{1+cos(x)}$
7. Apply the initial condition $y(0) = 1, 1 = \frac{C}{1+cos(0)} ↭ 1 = \frac{C}{2} ↭[\text{Solving for C}] C = 2$.
8. Final solution. Substitute C = 2 back into the general solution. The particular solution to the differential equation, satisfying y(0) = 1, is: $\boxed{y = \frac{x^2+2}{1+cos(x)}}$

• Solve $\frac{dy}{dx}+\frac{3y}{x} = \frac{e^x}{x^3}$

1. Write the differential equation in standard form. It is already in standard form where $p(x) = \frac{3}{x}, q(x) = \frac{e^x}{x^3}$
2. Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{3}{x}} = e^{3ln(x)} = e^{ln(x^3)} = x^3$ where x > 0 (to avoid division by zero and ensure μ(x) is defined).
3. Multiplying both sides of the differential equation by the integrating factor μ(x): $x^3\frac{dy}{dx} +3x^2y = e^x.$
4. By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $(x^3y)' = e^x$
5. Integrate both sides of the equation with respect to x: $x^3y = \int e^xdx = e^x + C$ where C is the constant of integration.
6. Solve for y, $\boxed{y = \frac{e^x + C}{x^3} = \frac{e^x}{x^3} + \frac{C}{x^3}}$ where C is an arbitrary constant determined by initial conditions (if any are provided).

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