Separable equations can be solved by moving all the $y$'s to one side. An equation like $y' + 2y = 4$ also separates, but $y' + 2y = e^{t}$ does not. Linear first-order equations need one new idea: multiply by a well-chosen function so that the left side becomes the derivative of a product.
Key idea
Write the equation in standard form $y' + p(t)y = g(t)$. Multiply both sides by the integrating factor $\mu(t) = e^{\int p(t)\,dt}$. The left side becomes $(\mu y)'$, so $$\mu y = \int \mu g\,dt + C,$$ $$y = \frac{1}{\mu}\left(\int \mu g\,dt + C\right).$$
Why it works
We want $\mu y' + \mu p y$ to equal $(\mu y)' = \mu y' + \mu' y$. Comparing the two, we need $\mu' = p\mu$, a separable equation whose solution is $\mu = e^{\int p\,dt}$. Any antiderivative of $p$ works, because a different constant only multiplies $\mu$ by a constant, which cancels at the end.
- Divide by the coefficient of $y'$ to reach standard form, and read off $p(t)$.
- Compute $\mu = e^{\int p\,dt}$ (no constant needed).
- Multiply both sides by $\mu$ and write the left side as $(\mu y)'$.
- Integrate both sides (with $+C$), divide by $\mu$, then use the initial condition.
Worked example: Constant coefficients
Solve $y' + 2y = 4$, $y(0) = 5$.
- $p = 2$, so $\mu = e^{2t}$.
- Multiply: $e^{2t}y' + 2e^{2t}y = (e^{2t}y)'$ $= 4e^{2t}$.
- Integrate: $e^{2t}y = 2e^{2t} + C$, so $y = 2 + Ce^{-2t}$.
- $y(0) = 2 + C = 5$ gives $C = 3$.
Answer: $y = 2 + 3e^{-2t}$. The 2 is the equilibrium; the $3e^{-2t}$ part dies out.
Worked example: Not yet in standard form
Solve $ty' + 2y = 6t$, $y(1) = 4$, for $t > 0$.
- Divide by $t$: $y' + \frac{2}{t}y = 6$, so $p = \frac{2}{t}$.
- $\mu = e^{2\ln t} = t^{2}$.
- Multiply: $(t^{2}y)' = 6t^{2}$, so $t^{2}y = 2t^{3} + C$ and $y = 2t + \frac{C}{t^{2}}$.
- $y(1) = 2 + C = 4$ gives $C = 2$.
Answer: $y = 2t + \frac{2}{t^{2}}$
Worked example: When the forcing matches the decay
Solve $y' - y = e^{t}$, $y(0) = -1$.
- $p = -1$, so $\mu = e^{-t}$.
- Multiply: $(e^{-t}y)' = e^{-t}e^{t} = 1$.
- Integrate: $e^{-t}y = t + C$, so $y = (t + C)e^{t}$.
- $y(0) = C = -1$.
Answer: $y = (t - 1)e^{t}$. The integral produced a factor $t$, a first sign of resonance.
Common mistake
Two slips are common. (1) Computing $\mu$ from the coefficient of $y$ before dividing by the coefficient of $y'$: for $ty' + 2y = 6t$, using $p = 2$ instead of $\frac{2}{t}$. (2) Multiplying only the left side by $\mu$. Check your $\mu$: after multiplying, the left side must be exactly $(\mu y)'$; expand it and compare.
The general solution of a linear equation always has the shape particular solution + $C$ times $e^{-\int p\,dt}$. The second part solves $y' + py = 0$. This split returns, in a bigger form, for second-order equations.