Turn a first-order equation into a product derivative

Try

The equation y′+y=t combines state-dependent loss with a time-dependent input. Trying to move all y terms to one side does not separate it into a product of one time function and one state function. Instead, look for a multiplier that turns the left side into a derivative of a product. Recall that (μy)′=μy′+μ′y.

For this example, multiplying by e^t works because its derivative is itself. Write out both sides of (e^t y)′=e^t(y′+y) before using the shortcut.

Predict

If the coefficient of y is p(t), what equation should the multiplier μ satisfy? Match the product-rule coefficients to predict μ′=pμ. Also predict why the multiplier should remain nonzero on the interval where we plan to divide by it at the end.

Experiment

Solve y′+y=t with y(0)=0. Multiplication gives (e^t y)′=te^t. Integration by parts yields ∫te^t dt=(t−1)e^t+C. Divide by e^t to get y=t−1+Ce^(−t), then use the initial condition to obtain C=1.

Check	Calculation	Result
Initial value	0−1+1	0
Derivative	1−e^(−t)	y′
Equation	1−e^(−t)+t−1+e^(−t)	t

Perform the same three checks after deliberately changing the sign of the exponential term. Identify which check catches the error. A correct-looking long-term trend can coexist with incorrect initial data.

Understand

For the normalized linear equation y′+p(t)y=q(t), choose μ(t)=exp(∫p(t)dt). Then μ′=pμ and (μy)′=μq. Integrating produces μy=∫μq dt+C. The arbitrary constant in the antiderivative defining μ can be omitted because any nonzero constant multiple works equally well.

On an interval where p and q are continuous, this procedure supplies the unique solution for an initial value in that interval. A definite-integral version automatically incorporates the initial condition: μ(t)y(t)=μ(t₀)y₀+∫ from t₀ to t of μ(s)q(s)ds. The variable s is only an integration variable; the upper endpoint t determines the evolving result.

For y′+y/t=2 on t>0, choose μ=t. Thus (ty)′=2t, giving ty=t²+C and y=t+C/t. With y(1)=3, C=2, so y(2)=3. The interval restriction is part of the solution, not a cosmetic footnote.

Consider the unnormalized equation t y′+y=t². Dividing by t excludes zero. At zero the original equation itself requires y(0)=0. A formula derived on a punctured interval does not automatically establish differentiability or satisfaction at the missing point. In this case (ty)′=t² gives y=t²/3+C/t away from zero; only C=0 has a differentiable extension through zero, and that extension must still be checked directly.

Apply

Compare a constant input y′+2y=6 with the time-varying input above. Its equilibrium is three. With y(0)=1, the initial slope is four and the solution is 3−2e^(−2t). Explain which term carries the initial deviation and which term represents the persistent input response.

For every integrating-factor calculation, write the normalized equation, the allowed interval, the product derivative, and the final verification. A coefficient with a singularity is a signal to analyze domains before manipulating symbols. Complete the practice by explaining the product rule rather than memorizing the multiplier alone.

Continue

The next lesson derives a linear equation from a material balance. Choosing the correct amount and volume variables is as important as solving the resulting equation accurately.