Second Order Differential Equation Basics
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Direct answer
For y″ + py′ + qy = 0 with constant coefficients, put y = e^{mx}; the trial survives only when m solves the auxiliary equation m² + pm + q = 0, and the nature of the roots writes the general solution by itself. Distinct real roots m1, m2 give y = c1 e^{m1 x} + c2 e^{m2 x}; a repeated root gives (c1 + c2x)e^{mx}; complex roots α ± iβ give e^{αx}(c1 cos βx + c2 sin βx). With a forcing term R(x), the complete solution is CF + PI — the complementary function plus one particular integral from an informed trial. Two initial conditions then pin c1 and c2 through simultaneous equations.
What you must remember
- Auxiliary equation: write m² + pm + q = 0 straight off y″ + py′ + qy = 0; no substitution ceremony is needed.
- Three root cases: distinct real → two exponentials; repeated → (c1 + c2 x)e^{mx}, the x-multiplication is mandatory; complex α ± iβ → e^{αx}(c1 cos βx + c2 sin βx).
- Complex case reading: the real part of the root drives the exponential, the imaginary part drives the oscillation frequency.
- CF + PI: the general solution of the non-homogeneous equation is one complementary function plus any one particular solution — never the PI alone.
- Trial discipline: R = polynomial → try same-degree polynomial; R = e^{kx} → try Ae^{kx} (multiply by x if k already solves the auxiliary equation); R = cos/sin → try A cos + B sin together.
- Superposition: for a sum of forcing terms, add the individual particular trials.
- Syllabus placement: the official NTA JEE Main syllabus names first order, first degree equations; constant-coefficient second order lives in Advanced archives and every serious coaching module — learn it after the first-order core is airtight.
One equation, two conditions
Solve y″ − 5y′ + 6y = 0 with y(0) = 1 and y′(0) = 0. The auxiliary equation is m² − 5m + 6 = 0, factoring as (m − 2)(m − 3) = 0 with roots 2 and 3 — distinct real, so y = c1 e^{2x} + c2 e^{3x}. Now the conditions: y(0) = c1 + c2 = 1, and y′ = 2c1 e^{2x} + 3c2 e^{3x} gives y′(0) = 2c1 + 3c2 = 0. Solving, multiply the first by 2 and subtract: c2 = −2, then c1 = 3. The solution is y = 3e^{2x} − 2e^{3x}. Verify both conditions in place: y(0) = 3 − 2 = 1 and y′(0) = 6 − 6 = 0. The architecture to notice: the differential equation chose the shape of the family (two exponentials), the initial conditions chose the member — and the two conditions consumed exactly the two constants a second-order equation provides. Had the auxiliary equation carried a double root, the same two conditions would fix c1 and c2 inside (c1 + c2 x)e^{mx} instead.
Where this sits in the syllabus
JEE Main's differential-equations quota comes overwhelmingly from separable and first-order first-degree homogeneous types, so second-order work there is mostly a safety net for the occasional memory-based item. JEE Advanced, by contrast, has historically set constant-coefficient second-order equations — including particular-integral trials — and the topic also underwrites the simple harmonic motion and damped-oscillation language of physics, which is why coaching modules teach it despite the syllabus's silence. The three habitual losses: the repeated-root case solved without the x, producing two proportional terms that masquerade as two constants; the trial Ae^{kx} used when k is already an auxiliary root, where the correct trial is Axe^{kx}; and sign errors assembling the simultaneous equations from initial conditions — differentiating the general solution and only then substituting x = 0, in that order, every time. Resonance-type questions (forcing at the natural frequency) are the Advanced flourish: recognise k as a root, upgrade the trial, and the extra x does the rest.
Frequently asked questions
What is the auxiliary equation of y″ + py′ + qy = 0?
m² + pm + q = 0, obtained by testing y = e^{mx}; its roots dictate the solution's form entirely.
What changes when the auxiliary equation has a repeated root?
The second solution gains a factor x: y = (c1 + c2 x)e^{mx} — without it the two constants collapse into one.
What form does the solution take for complex roots α ± iβ?
y = e^{αx}(c1 cos βx + c2 sin βx): exponential envelope from the real part, oscillation from the imaginary part.
Why is the general solution CF + PI?
Homogeneous solutions compose and scale; adding any single particular solution shifts the result onto the non-homogeneous equation — linearity does the bookkeeping.
Is this topic in the JEE Main syllabus?
Not by name — the NTA syllabus specifies first order, first degree; second-order constant-coefficient equations belong to Advanced-level preparation and physics-adjacent modules.