📐 Mathematics · Differential Equations · JEE

Solve: y'' - 4y = 0 with y(0) = 1, y'(0) = 0.

Answer: y = cosh(2x).

A y = cosh(2x)
B y = e^2x + e^-2x
C y = cos(2x)
D y = sinh(2x)

Correct answer: A. y = cosh(2x)

Explanation: Auxiliary: m² = 4, m = ±2. y = Ae^2x + Be^-2x. y(0)=A+B=1, y'(0)=2A-2B=0 so A=B=1/2. y = (e^2x+e^-2x)/2 = cosh(2x).

The general solution y=Ax² represents an entire FAMILY of curves, one for each value of the arbitrary constant A; a particular solution (fixed by an initial condition) selects exactly one curve from this family.

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Equations involving derivatives and their solutions

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Solve: y'' - 4y = 0 with y(0) = 1, y'(0) = 0.

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