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First order differential equations

Solving a differential equation determines a family of functions until a condition fixes the constant.

For a separable equation, move all factors involving y to one side and all factors involving x to the other, then integrate. Dividing by a factor of y can remove an equilibrium solution; check the original equation for those cases.

Apply an initial or boundary condition after finding the general solution. Substitute the result back into the original equation to check it. In a model, discuss whether the solution makes sense within its intended domain.

Worked example

Solve dy/dx = 2y with y(0) = 3.

  1. For y ≠ 0, integrate dy/y = 2 dx to get ln|y| = 2x + C.
  2. Write y = Ae^(2x).
  3. The initial condition gives A = 3.

Answer: y = 3e^(2x)

Revise first: Integration and exact area.

Forcing terms and repeated roots

A first-order linear equation y′ + py = q uses the integrating factor e^(∫p dx). Multiplying by it turns the left side into the derivative of a product. Keep the integration constant until you apply the initial condition.

For a constant-coefficient second-order homogeneous equation, substitute y = e^(rx) to obtain the auxiliary equation. Two distinct real roots give two exponential terms. A repeated root r instead gives (A + Bx)e^(rx); writing two copies of e^(rx) gives only one independent solution.

A forced equation needs a complementary function and a particular integral. A polynomial trial is useful for polynomial forcing, provided it is not already part of the complementary function. If it overlaps, multiply the trial by a sufficient power of x.

Solving y′ + 2y = 4x + 6, y(0) = 5

  1. Try y_p = ax + b. Then y_p′ + 2y_p = 2ax + a + 2b.
  2. Comparing coefficients gives a = 2 and 2 + 2b = 6, hence b = 2.
  3. The homogeneous solution is Ce^(−2x), so y = 2x + 2 + Ce^(−2x).
  4. The initial condition gives C = 3. Differentiate and substitute: y′ + 2y = 4x + 6, and y(0) = 5.
For y″ − 4y′ + 4y = 0, why is y = Ae^(2x) + Be^(2x) insufficient?

Both terms are multiples of the same function. The repeated root needs y = (A + Bx)e^(2x).

Take the idea further

With y(0) = 1 and y′(0) = 0 in that repeated-root equation, A = 1 and B = −2.

Explore the diagram and test a prediction

More advanced methods

Practise first order differential equations

Course mapping

These specification references show where the topic occurs. The questions cover only some parts of each topic.

Next practice: Hyperbolic functions.