The Central Idea
A first-order differential equation may be easy or difficult depending on how its variables and derivatives are arranged.
For example,
\[ \frac{dy}{dx}=xy \]can be reorganized so that all expressions involving \(y\) are placed with \(dy\), while all expressions involving \(x\) are placed with \(dx\).
By contrast,
\[ y'+2y=x \]is already arranged as a first-order linear equation.
The central skill is therefore not merely integration. It is classification followed by the method that matches the structure.
Two Major First-Order Forms
A separable equation can be written in the form
\[ \boxed{\frac{dy}{dx}=g(x)h(y)}. \]A linear first-order equation can be written in the form
\[ \boxed{y'+P(x)y=Q(x)}. \]Some equations belong to both classes, while others belong to only one.
Recognizing a Separable Equation
Consider
\[ \frac{dy}{dx}=x(1+y^2). \]The right side is a product of a function of \(x\), namely \(x\), and a function of \(y\), namely \(1+y^2\).
Therefore the equation is separable.
By contrast,
\[ \frac{dy}{dx}=x+y \]is not separable in this form because the variables occur as a sum rather than as a product that can be separated into an \(x\)-part and a \(y\)-part.
Separation of Variables
Suppose
\[ \frac{dy}{dx}=g(x)h(y). \]When \(h(y)\neq0\), divide by \(h(y)\):
\[ \frac{1}{h(y)}\frac{dy}{dx}=g(x). \]We then write the separated form as
\[ \boxed{\frac{1}{h(y)}\,dy=g(x)\,dx}. \]Now each side contains only one variable, so both sides can be integrated.
Why the Differential Notation Is Useful
The notation
\[ \frac{dy}{dx} \]behaves suggestively during separation, but the underlying justification is the chain rule and integration.
Starting from
\[ \frac{1}{h(y)}\frac{dy}{dx}=g(x), \]integrate with respect to \(x\):
\[ \int \frac{1}{h(y)}\frac{dy}{dx}\,dx = \int g(x)\,dx. \]The substitution \(u=y(x)\) turns the left side into an integral with respect to \(y\).
Thus the familiar differential notation accurately reflects the structure of the calculation.
The Basic Separable Workflow
- Write the equation as \(y'=g(x)h(y)\) when possible.
- Identify any equilibrium solutions before dividing by a factor involving \(y\).
- Move the \(y\)-dependent factor with \(dy\).
- Move the \(x\)-dependent factor with \(dx\).
- Integrate both sides.
- Combine constants of integration into a single constant.
- Solve explicitly for \(y\) when practical.
- Apply any initial condition.
- State the appropriate solution interval.
- Differentiate and substitute to verify.
Example: The Simplest Separable Equation
Solve
\[ \frac{dy}{dx}=xy. \]For nonzero \(y\), separate:
\[ \frac{1}{y}\,dy=x\,dx. \]Integrate:
\[ \int \frac{1}{y}\,dy = \int x\,dx. \] \[ \ln|y|=\frac{x^2}{2}+C. \]Exponentiate:
\[ |y|=e^C e^{x^2/2}. \]Absorb the sign and positive constant into a new arbitrary constant \(A\):
\[ \boxed{y=Ae^{x^2/2}}. \]The value \(A=0\) also includes the equilibrium solution \(y=0\).
Combining Constants Efficiently
After integration we often obtain one constant from each side:
\[ F(y)+C_1=G(x)+C_2. \]Rearranging gives
\[ F(y)=G(x)+(C_2-C_1). \]Since the difference of two arbitrary constants is still an arbitrary constant, write
\[ \boxed{F(y)=G(x)+C}. \]There is usually no benefit to carrying several unrelated constants through the solution.
Example: A Power of the Dependent Variable
Solve
\[ \frac{dy}{dx}=x y^2. \]Separate:
\[ y^{-2}\,dy=x\,dx. \]Integrate:
\[ \int y^{-2}\,dy = \int x\,dx. \] \[ -\frac{1}{y}=\frac{x^2}{2}+C. \]Solve for \(y\):
\[ y=-\frac{1}{x^2/2+C}. \]Multiplying numerator and denominator by \(2\) and renaming the constant gives a convenient equivalent form:
\[ \boxed{y=-\frac{2}{x^2+C}}. \]Do Not Lose Equilibrium Solutions
In the equation
\[ y'=xy^2, \]separation divides by \(y^2\).
But \(y=0\) makes that division invalid.
Check the original differential equation:
\[ y'=0, \qquad xy^2=0. \]Therefore,
\[ \boxed{y=0} \]is also a solution and must be recorded separately if it is not already included in the general family.
Example: A Separable Initial-Value Problem
Solve
\[ \frac{dy}{dx}=xy, \qquad y(0)=3. \]From the previous calculation,
\[ y=Ae^{x^2/2}. \]Apply the initial condition:
\[ 3=Ae^0. \]Thus,
\[ A=3. \]The particular solution is
\[ \boxed{y=3e^{x^2/2}}. \]Using the Initial Condition Early
An initial condition may be applied after solving explicitly, or sometimes earlier while the solution is still implicit.
Suppose
\[ \ln|y|=\frac{x^2}{2}+C \]and \(y(0)=3\).
Substitute immediately:
\[ \ln 3=C. \]Then
\[ \ln|y|=\frac{x^2}{2}+\ln3. \]Exponentiating leads to the same result,
\[ y=3e^{x^2/2}. \]Either approach is valid when the domain and sign information are handled correctly.
Example: An Implicit Separable Solution
Solve
\[ \frac{dy}{dx}=\frac{x}{y}. \]Separate:
\[ y\,dy=x\,dx. \]Integrate:
\[ \frac{y^2}{2}=\frac{x^2}{2}+C. \]Multiply by \(2\):
\[ y^2=x^2+C. \]The implicit form
\[ \boxed{y^2-x^2=C} \]may be more useful than splitting the solution into the two explicit branches
\[ y=\pm\sqrt{x^2+C}. \]Absolute Values in Logarithmic Integrals
The antiderivative
\[ \int \frac{1}{y}\,dy \]is
\[ \boxed{\ln|y|+C}, \]not merely \(\ln y+C\).
The absolute value allows the formula to represent both positive and negative nonzero values of \(y\) on an appropriate interval.
Example: Trigonometric Separation
Solve
\[ \frac{dy}{dx}=x\cos y. \]When \(\cos y\neq0\), separate:
\[ \sec y\,dy=x\,dx. \]Integrate:
\[ \int \sec y\,dy = \int x\,dx. \] \[ \ln|\sec y+\tan y| = \frac{x^2}{2}+C. \]This implicit form is a valid solution relation. Solving explicitly for \(y\) is not always the most efficient final step.
Separable Models and Proportional Change
The model
\[ \frac{dP}{dt}=kP \]is separable:
\[ \frac{1}{P}\,dP=k\,dt. \]Integrating gives
\[ \ln|P|=kt+C. \]Exponentiating yields
\[ \boxed{P=Ae^{kt}}. \]This derivation is the mathematical foundation for exponential growth and decay models, which receive a dedicated lesson later in this section.
Recognizing a Linear First-Order Equation
A linear first-order equation can be written as
\[ \boxed{y'+P(x)y=Q(x)}. \]The function \(y\) and its derivative \(y'\) appear only to the first power and are not multiplied together.
Examples include
\[ y'+3y=x, \] \[ y'-\frac{2}{x}y=x^2, \]and
\[ y'+(\sin x)y=e^x. \]What Makes an Equation Nonlinear?
Each of the following is nonlinear:
\[ y'+y^2=x, \] \[ yy'=x, \] \[ y'+\sin y=x. \]The first contains \(y^2\), the second multiplies \(y\) and \(y'\), and the third places \(y\) inside a nonlinear function.
Linearity refers to how the unknown function and its derivatives appear, not to whether the graph of the solution is a straight line.
Put the Equation in Standard Linear Form First
The integrating-factor method expects the coefficient of \(y'\) to be \(1\).
Consider
\[ x y'+2y=x^3. \]Divide by \(x\), assuming \(x\neq0\):
\[ y'+\frac{2}{x}y=x^2. \]Now we can identify
\[ P(x)=\frac{2}{x} \]and
\[ Q(x)=x^2. \]The Integrating Factor
For
\[ y'+P(x)y=Q(x), \]define the integrating factor
\[ \boxed{\mu(x)=e^{\int P(x)\,dx}}. \]Multiplying the entire differential equation by \(\mu(x)\) makes the left side become the derivative of a product.
Why the Integrating Factor Works
Multiply
\[ y'+P(x)y=Q(x) \]by \(\mu(x)\):
\[ \mu y'+\mu P y=\mu Q. \]We want the left side to equal
\[ \frac{d}{dx}(\mu y). \]By the product rule,
\[ \frac{d}{dx}(\mu y)=\mu y'+\mu' y. \]Thus we need
\[ \mu'=P\mu. \]The function
\[ \mu=e^{\int P(x)\,dx} \]satisfies exactly this requirement.
The Linear-Equation Workflow
- Rewrite the equation as \(y'+P(x)y=Q(x)\).
- Identify \(P(x)\) and \(Q(x)\).
- Compute \(\mu(x)=e^{\int P(x)\,dx}\).
- Multiply every term by \(\mu(x)\).
- Recognize the left side as \((\mu y)'\).
- Integrate both sides.
- Solve for \(y\).
- Apply any initial condition.
- State the appropriate interval.
- Verify in the original equation.
Example: Constant-Coefficient Linear Equation
Solve
\[ y'+2y=e^x. \]Here,
\[ P(x)=2. \]Therefore,
\[ \mu(x)=e^{\int2\,dx}=e^{2x}. \]Multiply every term by \(e^{2x}\):
\[ e^{2x}y'+2e^{2x}y=e^{3x}. \]The left side is
\[ \frac{d}{dx}(e^{2x}y). \]Thus,
\[ \frac{d}{dx}(e^{2x}y)=e^{3x}. \]Integrate:
\[ e^{2x}y=\frac{1}{3}e^{3x}+C. \]Divide by \(e^{2x}\):
\[ \boxed{y=\frac{1}{3}e^x+Ce^{-2x}}. \]Checking the Linear Solution
Let
\[ y=\frac{1}{3}e^x+Ce^{-2x}. \]Differentiate:
\[ y'=\frac{1}{3}e^x-2Ce^{-2x}. \]Then
\[ y'+2y = \frac{1}{3}e^x-2Ce^{-2x} +\frac{2}{3}e^x+2Ce^{-2x}. \]The terms containing \(C\) cancel:
\[ y'+2y=e^x. \]The differential equation is satisfied.
Example: A Linear Initial-Value Problem
Solve
\[ y'+2y=e^x, \qquad y(0)=4. \]The general solution is
\[ y=\frac{1}{3}e^x+Ce^{-2x}. \]Apply the initial condition:
\[ 4=\frac{1}{3}+C. \]Therefore,
\[ C=\frac{11}{3}. \]The particular solution is
\[ \boxed{ y=\frac{1}{3}e^x+\frac{11}{3}e^{-2x} }. \]Example: Variable Coefficients
Solve
\[ xy'+y=x^2, \qquad x>0. \]Divide by \(x\):
\[ y'+\frac{1}{x}y=x. \]Here,
\[ P(x)=\frac{1}{x}. \]The integrating factor is
\[ \mu(x)=e^{\int 1/x\,dx}. \]Since \(x>0\),
\[ \mu(x)=e^{\ln x}=x. \]Multiply through by \(x\):
\[ xy'+y=x^2. \]The left side is already
\[ \frac{d}{dx}(xy). \]Thus,
\[ (xy)'=x^2. \]Integrate:
\[ xy=\frac{x^3}{3}+C. \]Divide by \(x\):
\[ \boxed{y=\frac{x^2}{3}+\frac{C}{x}}, \qquad x>0. \]Intervals Matter in Linear Equations
In the previous example, dividing by \(x\) required
\[ x\neq0. \]Therefore the standard linear form is valid on intervals that do not cross zero, such as
\[ (0,\infty) \]or
\[ (-\infty,0). \]The integrating factor and the final solution should be interpreted on one such interval at a time.
The Constant in the Integrating Factor Is Unnecessary
Suppose
\[ \int P(x)\,dx=G(x)+C_0. \]Then
\[ e^{G(x)+C_0}=e^{C_0}e^{G(x)}. \]The factor \(e^{C_0}\) is a nonzero constant multiplier. Multiplying an integrating factor by a nonzero constant does not change its effectiveness.
Therefore we normally write simply
\[ \boxed{\mu(x)=e^{\int P(x)\,dx}} \]without adding an arbitrary constant inside the exponent.
Example: Integrating Factor with a Polynomial Coefficient
Solve
\[ y'+2xy=2x. \]Here,
\[ P(x)=2x. \]Therefore,
\[ \mu(x)=e^{\int2x\,dx}=e^{x^2}. \]Multiply through:
\[ e^{x^2}y'+2xe^{x^2}y=2xe^{x^2}. \]Recognize the product derivative:
\[ \frac{d}{dx}(e^{x^2}y)=2xe^{x^2}. \]Integrate:
\[ e^{x^2}y=e^{x^2}+C. \]Therefore,
\[ \boxed{y=1+Ce^{-x^2}}. \]Homogeneous Linear First-Order Equations
A linear equation of the form
\[ y'+P(x)y=0 \]is called homogeneous in the first-order linear sense.
It is both linear and separable:
\[ y'=-P(x)y. \]Separating gives
\[ \frac{1}{y}\,dy=-P(x)\,dx. \]Therefore,
\[ \ln|y|=-\int P(x)\,dx+C. \]Hence,
\[ \boxed{y=Ce^{-\int P(x)\,dx}}. \]The Homogeneous Part of a Linear Solution
For the nonhomogeneous equation
\[ y'+P(x)y=Q(x), \]the general solution is often viewed as
\[ \boxed{y=y_h+y_p}, \]where \(y_h\) solves the homogeneous equation
\[ y'+P(x)y=0 \]and \(y_p\) is one particular solution of the full equation.
This viewpoint becomes especially important for second-order equations and systems.
Some Equations Are Both Separable and Linear
Consider
\[ y'=3y. \]It is separable because
\[ \frac{1}{y}\,dy=3\,dx. \]It is also linear because it can be written
\[ y'-3y=0. \]Both methods produce
\[ \boxed{y=Ce^{3x}}. \]When more than one method applies, choose the method that makes the structure clearest and the computation shortest.
Some Linear Equations Are Not Separable
Consider
\[ y'+y=x. \]Rewriting gives
\[ y'=x-y. \]The right side cannot be written as a product \(g(x)h(y)\) in the ordinary separable form.
However, the equation is linear, so an integrating factor applies.
Some Separable Equations Are Not Linear
Consider
\[ y'=xy^2. \]This equation is separable:
\[ y^{-2}\,dy=x\,dx. \]But it is not linear because \(y\) appears squared.
Classification prevents us from forcing an equation into an inappropriate method.
A Classification Table
| Equation | Separable? | Linear? |
|---|---|---|
| \(y'=xy\) | Yes | Yes |
| \(y'=xy^2\) | Yes | No |
| \(y'+y=x\) | No | Yes |
| \(y'=x+y^2\) | No | No |
The table emphasizes that separability and linearity are different structural properties.
Verification by Differentiation
After obtaining a candidate solution, differentiate it and substitute into the original equation.
For example, suppose
\[ y=1+Ce^{-x^2}. \]Then
\[ y'=-2xCe^{-x^2}. \]Evaluate
\[ y'+2xy. \]We obtain
\[ -2xCe^{-x^2}+2x(1+Ce^{-x^2})=2x. \]Therefore the solution satisfies
\[ y'+2xy=2x. \]Initial Conditions Select One Curve
A first-order general solution typically contains one arbitrary constant.
An initial condition such as
\[ y(x_0)=y_0 \]usually determines that constant and selects one curve from the family.
This reflects a general principle: one first-order differential equation normally requires one independent initial condition to determine a unique member of its solution family when the usual existence and uniqueness conditions hold.
Common Error: Separating a Sum
From
\[ y'=x+y, \]it is not valid to write
\[ \frac{dy}{y}=\frac{dx}{x}. \]The right side \(x+y\) is a sum, not a product of an \(x\)-function and a \(y\)-function.
Separation of variables requires genuine multiplicative separation after algebraic rearrangement.
Common Error: Dividing Away a Solution
Suppose
\[ y'=x y(y-2). \]Dividing by \(y(y-2)\) excludes the cases
\[ y=0 \]and
\[ y=2. \]Both should be checked in the original equation before division.
In fact, both are equilibrium solutions.
Common Error: Forgetting Absolute Values
The integral
\[ \int \frac{1}{y}\,dy \]gives
\[ \ln|y|+C. \]Omitting the absolute value can incorrectly discard negative solution branches before the arbitrary constant is interpreted.
Common Error: Forgetting to Normalize a Linear Equation
For
\[ 2y'+4y=x, \]the standard linear form is obtained by dividing by \(2\):
\[ y'+2y=\frac{x}{2}. \]Therefore the coefficient used in the integrating factor is
\[ P(x)=2, \]not \(4\).
Common Error: Multiplying Only Part of the Equation
When an integrating factor \(\mu(x)\) is used, it must multiply every term in the equation.
From
\[ y'+P(x)y=Q(x), \]the correct multiplication is
\[ \mu y'+\mu P y=\mu Q. \]Multiplying only the left side changes the equation and destroys equivalence.
Common Error: Missing the Product Derivative
After multiplying by the integrating factor, do not integrate the two left-hand terms separately unless necessary.
The entire purpose of the method is to recognize
\[ \mu y'+\mu' y = (\mu y)'. \]This product derivative is the key structural simplification.
Common Error: Dropping the Constant After Integration
Once
\[ (\mu y)'=\mu Q \]is integrated, write
\[ \mu y=\int \mu Q\,dx+C. \]The constant is essential because it represents the homogeneous part of the general solution.
Common Error: Ignoring the Original Equation During Verification
Algebraic transformations can introduce domain restrictions or conceal excluded cases.
The final solution should be checked against the original differential equation, not only against a later rearranged form.
This is especially important when division by variables or functions has occurred.
A Decision Procedure
- Can the equation be written as \(y'=g(x)h(y)\)? If yes, separation is available.
- Can the equation be written as \(y'+P(x)y=Q(x)\)? If yes, the linear method is available.
- If both apply, choose the cleaner method.
- If neither applies, look for another class such as exact, Bernoulli, homogeneous-substitution, or numerical treatment.
- Before dividing by a function of \(y\), identify possible constant solutions.
- After solving, use initial data and verify the result.
Example: Full Separable Workflow
Solve
\[ \frac{dy}{dx}=2xy, \qquad y(0)=5. \]The equation is separable.
For nonzero \(y\), separate:
\[ \frac{1}{y}\,dy=2x\,dx. \]Integrate:
\[ \ln|y|=x^2+C. \]Exponentiate:
\[ y=Ae^{x^2}. \]Apply \(y(0)=5\):
\[ 5=A. \]Therefore,
\[ \boxed{y=5e^{x^2}}. \]Verify:
\[ y'=10xe^{x^2} \]and
\[ 2xy=2x(5e^{x^2})=10xe^{x^2}. \]The solution satisfies both the equation and the initial condition.
Example: Full Linear Workflow
Solve
\[ y'-y=x, \qquad y(0)=2. \]This is linear with
\[ P(x)=-1, \qquad Q(x)=x. \]The integrating factor is
\[ \mu(x)=e^{\int -1\,dx}=e^{-x}. \]Multiply through:
\[ e^{-x}y'-e^{-x}y=xe^{-x}. \]The left side is
\[ (e^{-x}y)'. \]Therefore,
\[ (e^{-x}y)'=xe^{-x}. \]Integrate the right side by parts:
\[ \int xe^{-x}\,dx=-(x+1)e^{-x}+C. \]Thus,
\[ e^{-x}y=-(x+1)e^{-x}+C. \]Multiply by \(e^x\):
\[ y=-(x+1)+Ce^x. \]Apply \(y(0)=2\):
\[ 2=-1+C. \]Therefore,
\[ C=3. \]The particular solution is
\[ \boxed{y=3e^x-x-1}. \]Practice 1: Classify
Determine whether the equation is separable, linear, both, or neither:
\[ y'=x y. \]Practice 2: Classify
Determine whether the equation is separable, linear, both, or neither:
\[ y'+3y=x^2. \]Practice 3: Classify
Determine whether the equation is separable, linear, both, or neither:
\[ y'=x y^3. \]Practice 4: Separable Equation
Solve
\[ y'=3xy. \]Practice 5: Separable Initial-Value Problem
Solve
\[ y'=2xy, \qquad y(0)=4. \]Practice 6: Do Not Lose the Equilibrium
Solve
\[ y'=x y^2 \]and identify any solution that could be lost during separation.
Practice 7: Linear Equation
Solve
\[ y'+y=1. \]Practice 8: Linear Initial-Value Problem
Solve
\[ y'+y=1, \qquad y(0)=3. \]Practice 9: Standard Form
Rewrite in standard linear form and identify \(P(x)\) and \(Q(x)\):
\[ 2x y'+4y=x^3, \qquad x\neq0. \]Practice 10: Find the Integrating Factor
Find an integrating factor for
\[ y'+\frac{3}{x}y=x^2, \qquad x>0. \]Practice 11: Solve a Variable-Coefficient Linear Equation
Solve
\[ y'+\frac{1}{x}y=x, \qquad x>0. \]Practice 12: Verify
Verify that
\[ y=1+Ce^{-x} \]solves
\[ y'+y=1. \]Practice Answers
Practice 1
The equation
\[ y'=xy \]is separable because the right side is a product of a function of \(x\) and a function of \(y\). It is also linear because it can be written
\[ y'-xy=0. \]Therefore,
\[ \boxed{\text{both separable and linear}}. \]Practice 2
The equation
\[ y'+3y=x^2 \]is already in standard linear form. It is not separable in the ordinary product form.
\[ \boxed{\text{linear only}}. \]Practice 3
The equation
\[ y'=xy^3 \]is separable, but \(y^3\) makes it nonlinear.
\[ \boxed{\text{separable only}}. \]Practice 4
Separate:
\[ \frac{1}{y}\,dy=3x\,dx. \]Integrate:
\[ \ln|y|=\frac{3}{2}x^2+C. \]Therefore,
\[ \boxed{y=Ce^{3x^2/2}}. \]Practice 5
The general solution is
\[ y=Ce^{x^2}. \]Apply \(y(0)=4\):
\[ 4=C. \]Thus,
\[ \boxed{y=4e^{x^2}}. \]Practice 6
For nonzero \(y\), separate:
\[ y^{-2}\,dy=x\,dx. \]Integrate:
\[ -\frac{1}{y}=\frac{x^2}{2}+C. \]Therefore a nonzero solution family is
\[ \boxed{y=-\frac{2}{x^2+C}}. \]Division by \(y^2\) excludes \(y=0\), but direct substitution shows that
\[ \boxed{y=0} \]is also a solution.
Practice 7
The integrating factor is
\[ \mu=e^x. \]Multiply through:
\[ e^x y'+e^x y=e^x. \]Therefore,
\[ (e^x y)'=e^x. \]Integrate:
\[ e^x y=e^x+C. \]Thus,
\[ \boxed{y=1+Ce^{-x}}. \]Practice 8
From Practice 7,
\[ y=1+Ce^{-x}. \]Apply \(y(0)=3\):
\[ 3=1+C. \]Hence,
\[ C=2. \]Therefore,
\[ \boxed{y=1+2e^{-x}}. \]Practice 9
Divide by \(2x\):
\[ y'+\frac{2}{x}y=\frac{x^2}{2}. \]Therefore,
\[ \boxed{P(x)=\frac{2}{x}}, \qquad \boxed{Q(x)=\frac{x^2}{2}}. \]Practice 10
Since
\[ P(x)=\frac{3}{x}, \]and \(x>0\),
\[ \mu(x) =e^{\int 3/x\,dx} =e^{3\ln x} =x^3. \]Therefore,
\[ \boxed{\mu(x)=x^3}. \]Practice 11
The integrating factor is
\[ \mu(x)=e^{\int1/x\,dx}=x, \qquad x>0. \]Multiply through:
\[ xy'+y=x^2. \]Therefore,
\[ (xy)'=x^2. \]Integrate:
\[ xy=\frac{x^3}{3}+C. \]Thus,
\[ \boxed{y=\frac{x^2}{3}+\frac{C}{x}}, \qquad x>0. \]Practice 12
Differentiate
\[ y=1+Ce^{-x}: \] \[ y'=-Ce^{-x}. \]Then
\[ y'+y = -Ce^{-x}+1+Ce^{-x} =1. \]Therefore,
\[ \boxed{y=1+Ce^{-x}} \]satisfies the differential equation.
What to Remember
A separable first-order equation can be arranged so each variable appears on its own side:
\[ \boxed{ \frac{dy}{dx}=g(x)h(y) \quad\Longrightarrow\quad \frac{1}{h(y)}\,dy=g(x)\,dx }. \]Then integrate both sides.
A linear first-order equation has the standard form
\[ \boxed{y'+P(x)y=Q(x)}. \]Its integrating factor is
\[ \boxed{\mu(x)=e^{\int P(x)\,dx}}. \]Multiplication by \(\mu\) creates the product derivative
\[ \boxed{(\mu y)'=\mu Q}. \]- Classify before calculating.
- Check for equilibrium solutions before dividing by a function of \(y\).
- Use absolute values when integrating \(1/y\).
- Normalize a linear equation before computing the integrating factor.
- Multiply every term by the integrating factor.
- Recognize the product derivative created by the integrating factor.
- Apply initial conditions only after preserving all valid solution branches.
- Respect interval and domain restrictions.
- Verify the final result in the original differential equation.
Where We Go Next
Separation of variables and integrating factors solve two broad and important classes of first-order differential equations. Another class arises when two differential terms combine as the total differential of an underlying potential function.
The next lesson, Exact Equations, develops the condition
\[ M(x,y)\,dx+N(x,y)\,dy=0 \]and shows how partial derivatives reveal whether a function \(F(x,y)\) exists with
\[ dF=M\,dx+N\,dy. \]That method connects first-order differential equations directly to multivariable calculus and potential functions.