The differential equation of the family of curves $\mathrm{y}=e^{x}(\mathrm{~A} \cos x+\mathrm{B} \sin x)$,…

The differential equation of the family of curves $\mathrm{y}=e^{x}(\mathrm{~A} \cos x+\mathrm{B} \sin x)$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants is
  1. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)+2 \mathrm{y}=0$
  2. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}-2\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)-2 \mathrm{y}=0$
  3. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)-2 \mathrm{y}=0$
  4. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}-2\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)+2 \mathrm{y}=0$

Solution

Given $y=e^{x}(A \cos x+B \sin x)$...(1) $\begin{aligned} \frac{d y}{d x} &=e^{x}(-A \sin x+B \cos x)+e^{x}(A \cos x+B \sin x) \\ \frac{d^{2} y}{d x^{2}} &=e^{x}(-A \cos x-B \sin x)+e^{x}(-A \sin x+B \cos x)+e^{x}(-A \sin x+B \cos x) \\ &-e^{x}(A \cos x+B \sin x) \\ &=2 e^{x}(-A \sin x+B \cos x) \end{aligned}$ Thus we get $\begin{aligned} y &=e^{x}(A \cos x+B \sin x) \\ \frac{d y}{d x} &=e^{x}(-A \sin x+A \cos x+B \cos x+B \sin x) \end{aligned}$ $\begin{aligned} & \frac{d^{2} y}{d x^{2}}=e^{x}(-2 A \sin x+2 B \cos x) \\ \therefore & \frac{d^{2} y}{d x^{2}}-\frac{2 d y}{d x}+2 y \\ &=e^{x}[-2 A \sin x+2 B \cos x+2 A \sin x-2 A \cos x-2 B \cos x-2 B \sin x+2 A \cos x+2 B \sin x] \\ &=0 \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

Practice more Hyperbola questions on Aicharya