Solve a 2nd order linear DE: y'' - 5y' + 6y = 0 — characteristic equation method

Question

Solve the differential equation $y'' - 5y' + 6y = 0$ .

(JEE Advanced 2021, similar pattern)

Solution — Step by Step

For a linear DE with constant coefficients, we assume $y = e^{mx}$ , where $m$ is a constant to be determined.

Then: $y' = me^{mx}$ and $y'' = m^2 e^{mx}$ .

Substituting into $y'' - 5y' + 6y = 0$ :

m^2 e^{mx} - 5me^{mx} + 6e^{mx} = 0

Since $e^{mx} \neq 0$ , divide through:

m^2 - 5m + 6 = 0

This is the characteristic equation (also called the auxiliary equation).

m^2 - 5m + 6 = 0

(m - 2)(m - 3) = 0

m = 2 \text{ or } m = 3

We get two distinct real roots.

When the characteristic equation has two distinct real roots $m_1$ and $m_2$ , the general solution is:

y = C_1 e^{m_1 x} + C_2 e^{m_2 x}

\boxed{y = C_1 e^{2x} + C_2 e^{3x}}

where $C_1$ and $C_2$ are arbitrary constants determined by initial/boundary conditions.

Why This Works

The assumption $y = e^{mx}$ works because exponential functions have the special property that their derivatives are proportional to themselves. This transforms the differential equation into an algebraic equation in $m$ , which is much easier to solve.

The two linearly independent solutions $e^{2x}$ and $e^{3x}$ form a fundamental set — any solution of the DE can be written as a linear combination of these two. This is guaranteed by the theory of linear ODEs: a 2nd order linear ODE always has exactly 2 linearly independent solutions.

Alternative Method — The three cases

Depending on the discriminant of the characteristic equation $am^2 + bm + c = 0$ :

Discriminant	Roots	General Solution
$b^2 - 4ac > 0$	Two distinct real: $m_1, m_2$	$C_1 e^{m_1 x} + C_2 e^{m_2 x}$
$b^2 - 4ac = 0$	Repeated real: $m$	$(C_1 + C_2 x)e^{mx}$
$b^2 - 4ac < 0$	Complex: $\alpha \pm i\beta$	$e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

Our problem falls in case 1 (discriminant $= 25 - 24 = 1 > 0$ ).

For JEE Advanced, you must know all three cases. The repeated root case (case 2) is the trickiest — note the extra $x$ factor in $(C_1 + C_2 x)e^{mx}$ . The complex root case (case 3) gives oscillatory solutions, which appear in problems involving simple harmonic motion and LC circuits.

Common Mistake

When roots are repeated ( $m_1 = m_2 = m$ ), students write the general solution as $y = (C_1 + C_2)e^{mx} = Ce^{mx}$ . This has only ONE arbitrary constant, not two — so it cannot be the general solution of a 2nd order DE (which always needs two constants). The correct form is $(C_1 + C_2 x)e^{mx}$ — the factor of $x$ in front of $C_2$ is essential for linear independence.

Question

Solution — Step by Step

Why This Works

Alternative Method — The three cases

Common Mistake

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