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📐 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<sup>2x</sup> + e<sup>-2x</sup>
  • C y = cos(2x)
  • D y = sinh(2x)

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

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

Family of Solution Curves: y = Ax²Each value of the constant A gives ONE particular curve from the family

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.

Concept context

Equations involving derivatives and their solutions

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