Differential Equations · step 4 of 14

Initial value problems & particular solutions

An initial value problem pairs an ODE with enough conditions at a single point to fix every arbitrary constant — one per order. The essential discipline is order of operations: find the GENERAL solution first, with its constants, and only then substitute the conditions. Applying conditions midway almost always gives a wrong answer.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Solve for the general solution firstIt must contain n arbitrary constants for an nth-order equation.
Substitute the initial conditionsFor second order, you need both y(x₀) and y′(x₀), so differentiate the general solution before applying the second condition.
Solve the resulting simultaneous equationsTwo conditions give two linear equations in the two constants.
Verify by substitutionCheck the particular solution satisfies both the ODE and every condition.

Worked example

Solve y″ − y = 0 with y(0) = 2 and y′(0) = 0.

Characteristic equation r² − 1 = 0 gives r = ±1, so y = Ae^x + Be^(−x).
y(0) = 2: A + B = 2.
Differentiate: y′ = Ae^x − Be^(−x). Then y′(0) = 0 gives A − B = 0, so A = B.
Substituting: 2A = 2, so A = B = 1.

Answer. y = e^x + e^(−x) = 2cosh x. Check: y″ = e^x + e^(−x) = y ✓, y(0) = 2 ✓, y′(0) = 0 ✓.

Where marks get dropped

These are the specific errors that cost credit on initial value problems & particular solutions questions — QED's rubric penalises each of them separately.

Applying an initial condition before finding the full general solution, which loses part of the solution family.
Forgetting to differentiate before using a condition on y′. The second condition applies to the derivative of your general solution.
Supplying conditions at two different points. That is a BOUNDARY value problem, which may have no solution or infinitely many.

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Initial value problems & particular solutions — frequently asked questions

When is a solution guaranteed unique?

By the Picard–Lindelöf theorem, when f and ∂f/∂y are continuous near the initial point. Linear equations with continuous coefficients always satisfy this.

What is the difference from a boundary value problem?

An IVP gives all conditions at one point and behaves well. A BVP gives conditions at two points and can have zero, one or infinitely many solutions.

Can an IVP have no solution?

For a well-posed linear problem, no. Non-linear equations can fail — dy/dx = y² with y(0) = 1 blows up at x = 1, so the solution exists only locally.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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