Mastering Differential Equation Solving in Wolfram Alpha: Worked Examples, Part 2

Wolfram Alpha queries work best when they contain a complete equation and the conditions needed to identify a solution. A fragment such as “solve dy/dx” leaves the function unspecified. This lesson replaces those fragments with concrete problems you can check by hand.

Write the equation and its conditions

Identify the independent variable, unknown function, derivative order, and initial or boundary conditions. A second-order equation generally needs two independent conditions for a particular solution; giving only one may leave an arbitrary constant.

The official Wolfram Alpha differential-equation examples show complete equations and initial values. Inspect the interpreted input before using a result: a parsed equation that differs from your intended equation answers a different question.

Worked example 1: exponential decay

Consider y′(t) = −2y(t), with y(0) = 3. A complete query is y'(t) = -2 y(t), y(0) = 3.

Separate the variables: dy/y = −2 dt. Integrating gives ln|y| = −2t + C, and therefore y(t) = A e^−2t. The initial value sets A = 3, so y(t) = 3e^−2t.

Check both parts: its derivative is −6e^−2t, equal to −2y(t), and y(0) = 3. At t = 1, y ≈ 0.4060. The negative coefficient describes decay, not growth.

Worked example 2: a second-order initial-value problem

Solve y″ + y = 0 with y(0) = 2 and y′(0) = 1. The query y'' + y = 0, y(0)=2, y'(0)=1 appears in Wolfram’s official examples.

The general solution is A cos x + B sin x. Substituting x = 0 gives A = 2. Differentiating gives y′ = −A sin x + B cos x, so the second initial condition gives B = 1. Thus y = 2 cos x + sin x.

Differentiate twice to verify y″ = −y. Be careful to distinguish y′(0), a derivative value, from y(1), a function value at another point.

Worked example 3: understand an Euler step

For y′ = −2y, y(0) = 3, Euler’s method with h = 0.1 uses y_n+1 = y_n + h(−2y_n).

1. At t = 0.1: y₁ = 3 − 0.6 = 2.4.
2. At t = 0.2: y₂ = 2.4 − 0.48 = 1.92.

The exact value at t = 0.2 is 3e^−0.4 ≈ 2.0110. The numerical result is an approximation. A solver output should identify the interval and numerical method before you compare it with these steps.

Common input mistakes

• Missing the right-hand side of an equation.
• Switching between x and t without defining their roles.
• Omitting an initial value and expecting a unique curve.
• Using a numerical-method name without an equation or interval.
• Treating Wolfram Language functions such as NDSolve as interchangeable with Wolfram Alpha’s natural-language queries.

For complicated queries, start with the full equation, then add conditions. Use the official examples as syntax references rather than assuming every descriptive phrase is a supported command. Results and available detail can depend on the service’s interpretation and access level.

Practice problem

Solve y′ = 3y with y(0) = 2, then verify the differential equation and the initial condition.

Review the first Wolfram Alpha lesson for context, Introduction to Differential Equations for terminology, and Numerical Differentiation for estimating rates from samples.

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