ODE 2nd orderMedium

2nd order ODE: characteristic equation

Find the general solution of y'' − 5y' + 6y = 0.

Given data

y'' − 5y' + 6y = 0

Steps

0 / 3

Write the characteristic equation.
Write your working — the numerical result must appear in the text (logica)
Find λ₁ (the larger root).
Find λ₂ and write the general solution.

Full worked solution

Write the characteristic equation.
$\lambda^2 - 5\lambda + 6 = 0$
The characteristic equation is $\lambda^2 - 5\lambda + 6 = 0$ . Obtained by substituting $y = e^{\lambda x}$ into the ODE.
Find λ₁ (the larger root).
$\lambda_1 = 3$
$\lambda^2 - 5\lambda + 6 = (\lambda - 3)(\lambda - 2) = 0 \rightarrow \lambda_1 = \mathbf{3}$ and $\lambda_2 = 2$ . The larger root is $\lambda_1 = 3$ .
Find λ₂ and write the general solution.
$\lambda_2 = 2 \Rightarrow y = C_1 e^{3x}+C_2 e^{2x}$
$\lambda_2 = 2$ . For distinct real roots, the general solution is $y = C_1 e^{3x} + C_2 e^{2x}$ . Each root contributes an exponential term.

Result:y = C₁e^{3x} + C₂e^{2x}.

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