Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

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)=eP(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 My=Nx\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)=eP(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=Mdx+NdydF=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)=eP(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 My=Nx\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(4y)\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}.