First-order ODE basics: antiderivatives, constants, and checking

Key idea

A first-order ODE involves \(y(x)\) and its first derivative \(y'=\frac{dy}{dx}\). When the right side depends only on \(x\), the solution is an antiderivative: \[ \frac{dy}{dx}=f(x)\quad \Rightarrow \quad y=\int f(x)\,dx + C. \] The constant \(C\) appears because different solution curves can have the same derivative (they are vertical shifts of each other).

Solution check (the habit that saves points)

After you find \(y(x)\), differentiate it and substitute into the original ODE. If it matches, your solution is correct.

Worked example

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Summary

• If \(y'=f(x)\), integrate: \(y=\int f(x)\,dx + C\).
• Use initial conditions to determine \(C\), then check by differentiating.

Separable differential equations: separation of variables

Key idea

An ODE is separable if you can rewrite it as \[ \frac{dy}{dx}=f(x)\,g(y). \] Then you can separate variables: \[ \frac{1}{g(y)}\,dy = f(x)\,dx, \] integrate both sides, and include \(+C\).

Worked example

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Summary

• Separable form: \(y'=f(x)g(y)\) \(\Rightarrow\) \(\frac{1}{g(y)}dy=f(x)dx\).
• Integrate both sides and include a constant \(C\). Then solve for \(y\) if possible.

The exponential model \(y'=ky\): growth, decay, and matching solutions

Key idea

The differential equation \[ \frac{dy}{dx}=ky \] has general solution \[ y = Ce^{kx}. \] This model appears in exponential growth/decay and many “rate proportional to amount” situations.

Worked example

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Summary

• \(y'=ky\) \(\Rightarrow\) \(y=Ce^{kx}\).
• Always check by differentiating and substituting back into the ODE.

Linear first-order ODEs: standard form and integrating factor

Key idea

A linear first-order ODE has the form \[ y' + P(x)y = Q(x). \] The integrating factor is \[ \mu(x) = e^{\int P(x)\,dx}. \] Multiplying the ODE by \(\mu(x)\) makes the left side a product derivative: \[ \mu y' + \mu P y = (\mu y)'. \] Then integrate: \[ (\mu y)' = \mu Q \quad \Rightarrow \quad \mu y = \int \mu Q\,dx + C. \]

Worked example

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Summary

• Linear first-order: \(y' + P(x)y = Q(x)\).
• Integrating factor: \(\mu(x)=e^{\int P(x)\,dx}\), then \((\mu y)'=\mu Q\).

Recognize the ODE type: separable vs linear vs “not separable”

Key idea

Before solving, do a fast “type check”:

• Direct integration: if \(y'=f(x)\) only.
• Separable: if you can rearrange into \(g(y)\,dy=f(x)\,dx\).
• Linear: if it fits \(y' + P(x)y = Q(x)\) (even if it is also separable sometimes).

This 10-second decision is the biggest speed boost in first-order ODE practice.

Worked example

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Summary

• Classification comes first: it tells you the fastest correct method.
• \(y'=x+y\) is linear, but not separable in the simple \(f(x)g(y)\) product form.

Initial value problems: turning a family into one curve

Key idea

A general solution contains an arbitrary constant \(C\). An initial condition like \(y(x_0)=y_0\) chooses the unique member of the family (when it exists).

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Summary

• General solution \(\Rightarrow\) contains \(C\). Initial condition \(\Rightarrow\) find \(C\).
• Always check by differentiating after you apply the initial condition.

Why first-order ODEs matter (and a final check)

Where first-order ODEs show up

• Growth/decay: \(y'=ky\) models populations and radioactive decay.
• Motion with constant acceleration: velocity and position come from integrating derivatives.
• Electrical circuits: first-order linear ODEs in simple RC circuits (current/voltage relationships).
• Cooling: Newton’s law of cooling uses \(T' = -k(T-T_{\text{env}})\).

Worked example: verify a solution by substitution

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Final recap

• General vs particular: solve first, then use the initial condition to find \(C\).
• Separable ODEs: rearrange into \(g(y)\,dy=f(x)\,dx\), integrate, and solve for \(y\) if possible.
• Linear ODEs: \(y'+P(x)y=Q(x)\), integrating factor \(\mu=e^{\int P\,dx}\), then \((\mu y)'=\mu Q\).
• Always check: differentiate your solution and substitute back into the original ODE.