Linear ODE y' = ay + b: the one-formula solution

The theorem

For the ODE $y' = ay + b$ ($a \neq 0$, constants), solutions are:

$y(x) = C e^{ax} - \dfrac{b}{a}$

with $C \in \mathbb{R}$ a constant.

Break it into 2 pieces

General solution = homogeneous solution + constant particular solution.

• Homogeneous $y' = ay$ ⇒ $y_h = Ce^{ax}$
• Particular (constant $y_p$): solve $0 = a y_p + b$, so $y_p = -b/a$
• Total: $y = Ce^{ax} - b/a$

Special case: $b = 0$

If $b = 0$, equation becomes $y' = ay$ and $y = Ce^{ax}$. Exponential growth/decay.

Finding $C$ with an initial condition

A condition like "$y(0) = y_0$" or "$y(x_0) = y_0$" pins down $C$.

Example: solve $y' = 3y - 6$ with $y(0) = 4$.

• General: $y(x) = C e^{3x} + 2$.
• Condition: $4 = C + 2$, so $C = 2$.
• Unique: $y(x) = 2 e^{3x} + 2$.

Verification

Always plug back in: $y' = 6 e^{3x}$, and $3y - 6 = 6e^{3x} + 6 - 6 = 6e^{3x}$. ✓

Real applications

• Capacitor charging: $u' = -u/(RC) + E/(RC)$
• Newton’s cooling: $T' = -k(T - T_\text{env})$
• Malthusian growth ($b = 0$): $N' = kN$

Common mistakes

1. Confusing $y' = ay + b$ and $y' = ay - b$.
2. Forgetting the particular solution: $Ce^{ax}$ alone works only if $b = 0$.
3. Sign error on $-b/a$.

?À toi de jouer

Solve y' = 2y − 10 with y(0) = 3.

A.y(x) = −2e^{2x} + 5B.y(x) = 3e^{2x} − 10C.y(x) = e^{2x} + 5D.y(x) = 3e^{2x} + 5

Valider