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First-Order ODE Solution Methods Reference cheat sheet - grade 11-12

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Calculus Grade 11-12

First-Order ODE Solution Methods Reference Cheat Sheet

A printable reference covering separable equations, linear integrating factors, exact equations, slope fields, and equilibrium solutions for grades 11-12.

TopicsDerivativesIntegralsFunctionsGraphsCalculusStudy Guide
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First-order ordinary differential equations describe how a quantity changes when its rate of change depends on the variable, the quantity, or both. This cheat sheet helps students recognize common first-order ODE forms and choose an efficient solution method. It is useful for calculus, AP Calculus enrichment, and introductory differential equations practice.

The main goal is to connect the structure of an equation to the method that solves it.

Key Facts

  • A separable differential equation can be written as dydx=g(x)h(y)\frac{dy}{dx}=g(x)h(y), then solved by 1h(y) dy=g(x) dx\frac{1}{h(y)}\,dy=g(x)\,dx and integrating both sides.
  • A first-order linear differential equation has the form dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x).
  • The integrating factor for dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x) is μ(x)=e∫P(x) dx\mu(x)=e^{\int P(x)\,dx}.
  • After multiplying a linear equation by μ(x)\mu(x), the left side becomes ddx[μ(x)y]=μ(x)Q(x)\frac{d}{dx}[\mu(x)y]=\mu(x)Q(x).
  • An exact differential equation has the form M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0 and is exact when ∂M∂y=∂N∂x\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.
  • For an exact equation, the solution is F(x,y)=CF(x,y)=C, where Fx=MF_x=M and Fy=NF_y=N.
  • An equilibrium solution occurs when dydx=0\frac{dy}{dx}=0, so the solution is a constant function y=cy=c.
  • An initial condition such as y(x0)=y0y(x_0)=y_0 is used after finding the general solution to determine the constant CC.

Vocabulary

First-order ODE
A differential equation involving an unknown function and its first derivative, such as dydx=f(x,y)\frac{dy}{dx}=f(x,y).
Separable equation
A differential equation whose variables can be separated into the form A(y) dy=B(x) dxA(y)\,dy=B(x)\,dx before integration.
Linear equation
A first-order equation that can be written as dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x).
Integrating factor
A function μ(x)=e∫P(x) dx\mu(x)=e^{\int P(x)\,dx} used to turn a linear ODE into a product derivative.
Exact equation
An equation M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0 for which a potential function F(x,y)F(x,y) satisfies dF=M dx+N dydF=M\,dx+N\,dy.
Slope field
A graph showing small line segments with slope dydx=f(x,y)\frac{dy}{dx}=f(x,y) at many points to visualize solution curves.

Common Mistakes to Avoid

  • Separating variables incorrectly, because terms involving yy must stay with dydy and terms involving xx must stay with dxdx before integrating.
  • Forgetting the constant of integration, because solving ∫A(y) dy=∫B(x) dx\int A(y)\,dy=\int B(x)\,dx requires a constant CC on one side.
  • Using the wrong integrating factor, because μ(x)=e∫P(x) dx\mu(x)=e^{\int P(x)\,dx} only works after the equation is in the form dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x).
  • Calling an equation exact without checking, because exactness requires ∂M∂y=∂N∂x\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.
  • Applying the initial condition too early, because it should usually be used after finding the general solution or implicit solution.

Practice Questions

  1. 1 Solve the separable differential equation dydx=3x2y\frac{dy}{dx}=3x^2y with initial condition y(0)=2y(0)=2.
  2. 2 Solve the linear differential equation dydx+2y=ex\frac{dy}{dx}+2y=e^x.
  3. 3 Determine whether (2xy+3) dx+(x2+4y) dy=0(2xy+3)\,dx+(x^2+4y)\,dy=0 is exact, and if it is exact, find the implicit solution.
  4. 4 For the equation dydx=y(4−y)\frac{dy}{dx}=y(4-y), explain what the equilibrium solutions are and describe which one is stable using the sign of dydx\frac{dy}{dx}.

Understanding First-Order ODE Solution Methods Reference

Method choice depends on the algebra before any integration begins. A useful first step is to isolate the rate of change, then inspect which factors contain the dependent variable. If every term involving the dependent variable can be moved to one side while every term involving the independent variable goes to the other, separation is a natural choice.

This step can involve dividing by an expression that might equal zero. That creates an important risk.

Any constant value that makes a divided factor zero must be tested separately, since it may be a valid solution that disappears during the algebra. Students often lose equilibrium solutions this way.

The integrating factor method works because it deliberately creates a product rule. The original equation may contain a term involving the unknown function and a term involving its rate of change. Multiplying by the right function of the independent variable makes those pieces combine into the derivative of one product.

After integration, the unknown function is still inside that product, so it must be isolated carefully. Errors often happen when students forget to multiply every term by the integrating factor or use an antiderivative with the wrong sign.

It helps to differentiate the final answer and substitute it back into the original equation. This check is usually faster than redoing the whole problem.

Exact equations have a different idea behind them. They come from a hidden function whose total change is zero along a solution curve. Finding that hidden function is like rebuilding a landscape from information about its horizontal and vertical slopes.

Integrate one part with respect to its matching variable, then include an unknown function of the other variable. That extra function is necessary because differentiation can erase information. Comparing with the remaining part reveals it.

The equality test for exactness is not just a rule to memorize. It checks whether the two given slope pieces can belong to one consistent underlying function.

Slope fields give a visual test for algebraic work. Each small segment shows the direction a solution would take at that location. A proposed solution curve should follow those segments rather than cutting across them at a sharp angle.

Horizontal rows or curves in a slope field often signal equilibria. Nearby segments show whether solutions move toward an equilibrium or away from it. This is useful in population models, cooling problems, chemical mixing, and any setting with a steady level.

An initial condition selects one path from a family of possible paths. Keep track of where a formula is defined, since division by zero, logarithms, and roots can restrict the interval on which the selected solution is valid.