Exact & Linear Differential Equations

Exact & Linear Differential Equations

When separation of variables fails, we look for other structures. Two powerful methods are checking for exactness and transforming the equation into a linear form.

Exact Differential Equation

A differential equation of the form $M(x,y)dx + N(x,y)dy = 0$ is exact if the mixed partial derivatives are equal:

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Interact with the simulation below to explore integrating factors for exact and linear equations.

Linear DE

y' + \frac{2}{x}y = 4x

Find an integrating factor $\mu(x) = x^k$ to make it exact.

Select power $k$

k

0.0

✗ Not Exact

\mu(x) = x^{0}

\frac{d}{dx}\mu(x)

P(x)\mu(x)

To be exact, we need $\frac{d\mu}{dx} = P(x)\mu(x)$ . Match the blue curve to the dashed red curve.

Method for Exact DEs

If exact, there exists a potential function $F(x,y) = C$ such that:

\frac{\partial F}{\partial x} = M \quad \text{and} \quad \frac{\partial F}{\partial y} = N

Steps to Solve:

Integrate $M$ with respect to $x$ (treating $y$ as constant):
$F(x,y) = \int M \, dx + g(y)$
Differentiate this result with respect to $y$ and set it equal to $N$ :
$\frac{\partial}{\partial y} \left( \int M \, dx + g(y) \right) = N$
Solve for $g'(y)$ and integrate to find $g(y)$ .
The general solution is $F(x,y) = C$ .

Theory of Integrating Factors

An Integrating Factor (IF), denoted by $\mu(x,y)$ , is a function that, when multiplied by a non-exact differential equation $Mdx + Ndy = 0$ , turns it into an exact differential equation $(\mu M)dx + (\mu N)dy = 0$ .

This means the new equation must satisfy the exactness condition:

\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N)

Integrating Factors for Non-Exact DEs

If an equation is not exact ( $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ ), we can sometimes find an Integrating Factor (IF) that depends only on one variable, either $\mu(x)$ or $\mu(y)$ . We check this by taking the difference of the partials and dividing by $M$ or $N$ .

Case 1: IF depends only on $x$
Check the quantity: $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$ . If this expression simplifies to a function $f(x)$ dependent only on $x$ (no $y$ terms), then the integrating factor is $\mu(x) = e^{\int f(x)dx}$ .
Case 2: IF depends only on $y$
Check the quantity: $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$ . If this expression simplifies to a function $g(y)$ dependent only on $y$ (no $x$ terms), then the integrating factor is $\mu(y) = e^{\int g(y)dy}$ .

Multiplying the original DE by the corresponding $\mu$ will make it exact. You can then solve using the exact equation procedure.

Integrating Factor Visualizer

Linear First-Order Differential Equations

A first-order DE is linear if the dependent variable $y$ and its derivative $y'$ appear to the first power.

Standard Linear Form

\frac{dy}{dx} + P(x)y = Q(x)

Method of Integrating Factors (Linear)

To solve a linear DE:

Calculate IF: Ensure the equation is in the standard form. Find the integrating factor $\mu(x) = e^{\int P(x)dx}$ .
Multiply: Multiply the entire DE by $\mu(x)$ . The left side automatically becomes the derivative of a product:
$\frac{d}{dx}[\mu(x)y] = \mu(x)Q(x)$
Integrate: Integrate both sides with respect to $x$ :
$\mu(x)y = \int \mu(x)Q(x) \, dx + C$
Solve for y: Divide by $\mu(x)$ to obtain the explicit general solution.

Bernoulli's Equation

A Bernoulli Equation is a specific type of non-linear DE that can be systematically reduced to a linear one using a standard substitution.

Bernoulli Equation Form

\frac{dy}{dx} + P(x)y = Q(x)y^n

where $n$ is any real number and $n \neq 0, 1$ . (If $n=0$ or $n=1$ , the equation is already linear).

Bernoulli Substitution Method

Standardize: Write the equation in the Bernoulli form $y' + P(x)y = Q(x)y^n$ .
Divide: Divide the entire equation by $y^n$ to get: $y^{-n}y' + P(x)y^{1-n} = Q(x)$ .
Substitute: Introduce a new variable $v = y^{1-n}$ .
Differentiate: Differentiate $v$ with respect to $x$ : $v' = (1-n)y^{-n}y'$ . Note that this is proportional to the first term in step 2.
Transform: Substitute $v$ and $v'$ into the equation from step 2. You will obtain a standard linear DE in terms of $v$ and $x$ : $\frac{1}{1-n}v' + P(x)v = Q(x)$ .
Solve for v: Multiply by $(1-n)$ and solve this new linear equation using the Integrating Factor method.
Back-substitute: Replace $v$ with $y^{1-n}$ to find the solution in terms of $y$ .

Key Takeaways

Exact Equations rely on partial derivatives; verify $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . The solution is a potential surface $F(x,y)=C$ .
Integrating Factors (Non-Exact) can convert certain non-exact equations into exact ones by multiplying by $\mu(x)$ or $\mu(y)$ .
Linear Equations follow the standard form $y' + P(x)y = Q(x)$ and are solved using an Integrating Factor $\mu(x) = e^{\int P(x)dx}$ .
Bernoulli Equations are non-linear ( $y' + P(x)y = Q(x)y^n$ ) but can be transformed into linear ones via the substitution $v = y^{1-n}$ .
Always try to identify if an equation is Exact or Linear first, as these have reliable algorithmic solutions.

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