Worked example 1: Solve C' + 0.4 C = 8 with C(0) = 0. Interpret the equilibrium response.

integrating factor mu(t) = e^{0.4t}
The coefficient multiplying C determines the integrating factor through e^{integral p(t) dt}.
(e^{0.4t} C)' = 8 e^{0.4t}
This is the product-rule repair that makes the left side integrable in one step.
e^{0.4t} C = 20 e^{0.4t} + K
Integrating the right side adds a constant, just as in ordinary antiderivative work.
C(t) = 20 + K e^{-0.4t}
The solution naturally splits into a steady value plus a decaying transient.
C(t) = 20 (1 - e^{-0.4t})
The initial value forces K = -20. The equilibrium concentration is 20, and the transient only controls how fast the solution reaches it.

Worked example 2: Solve the RC-style model V' + 0.5 V = 4 with V(0) = 1.

steady value from 0 + 0.5 V = 4 is V_eq = 8
Setting the derivative to zero shows the long-term value the forcing tries to impose.
V(t) = 8 + K e^{-0.5t}
The integrating-factor method produces the same steady-plus-transient pattern as the infusion model.
V(t) = 8 - 7 e^{-0.5t}
The initial value only chooses the transient coefficient. It does not change the equilibrium set by the forcing balance.

Worked example 3: Recognize an exact equation in (2xy + 3) dx + (x^2 + 4y) dy = 0.

M_y = 2x and N_x = 2x
Matching cross-partial derivatives are the clue that the equation is exact.
Phi_x = 2xy + 3 gives Phi = x^2 y + 3x + g(y)
Integrate one coefficient and leave room for an unknown function of the other variable.
Phi_y = x^2 + g'(y) = x^2 + 4y
Matching the second coefficient fixes g'(y) = 4y, so g(y) = 2y^2.
x^2 y + 3x + 2y^2 = C
Exact equations are solved by a potential function rather than by an integrating factor.

Common mistake: choosing the integrating factor from the forcing term instead of from p(t). Another common mistake is treating the equilibrium value as the whole solution and forgetting the transient memory of the initial condition.

Differential equations