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Mathematics20234 marksDerivationContinuity, Differentiability and Differentiation

Prove the following.

If y = e^(ax) cos bx, then prove that (d²y)/(dx²) - 2a(dy)/(dx) + (a² + b²)y = 0

If $y = e^{ax} \cos bx$, then prove that $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$

Given: $y = e^{ax} \cos bx$

To show: $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$

Answer

Answer

$y = e^{ax}\cos bx$ $\frac{dy}{dx} = ae^{ax}\cos bx - be^{ax}\sin bx = e^{ax}(a\cos bx - b\sin bx)$ $\frac{d^2y}{dx^2} = ae^{ax}(a\cos bx - b\sin bx) + e^{ax}(-ab\sin bx - b^2\cos bx) = e^{ax}[(a^2-b^2)\cos bx - 2ab\sin bx]$ $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y = e^{ax}[(a^2-b^2)\cos bx - 2ab\sin bx - 2a^2\cos bx + 2ab\sin bx + (a^2+b^2)\cos bx]$ $= e^{ax}\cos bx\,[a^2 - b^2 - 2a^2 + a^2 + b^2] = 0$ Hence proved.
  1. Step 1: Given $y = e^{ax}\cos bx$.
  2. Step 2: Differentiating by the product rule: $\frac{dy}{dx} = e^{ax}(a\cos bx - b\sin bx)$.
  3. Step 3: Differentiating again by the product rule: $\frac{d^2y}{dx^2} = e^{ax}[(a^2-b^2)\cos bx - 2ab\sin bx]$.
  4. Step 4: Substituting in the left side: $e^{ax}[(a^2-b^2)\cos bx - 2ab\sin bx - 2a^2\cos bx + 2ab\sin bx + (a^2+b^2)\cos bx]$.
  5. Step 5: The $\sin bx$ terms cancel and the coefficient of $\cos bx$ is $a^2 - b^2 - 2a^2 + a^2 + b^2 = 0$.
  6. Step 6: Hence $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y = 0$.