Q118 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

If A and B are arbitrary constants, then the differential equation having $\mathrm{y}=\mathrm{Ae}^{\mathrm{x}}+\mathrm{B} \sin 2 \mathrm{x}$ as its general solution is

  • Correct answer
    A
    $\begin{aligned} &(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x}-4(\sin 2 x \\ &+\cos 2 x) y=0\end{aligned}$
  • B
    $\begin{aligned} &(\cos 2 x+\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x}-4(\sin \\ &2 x-\cos 2 x) y=0\end{aligned}$
  • C
    $\begin{aligned}(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} & +4(\sin 2 x \\ & +\cos 2 x) y=0\end{aligned}$
  • D
    $\begin{array}{r}(\sin 2 x-\cos 2 x) \frac{d^2 y}{d x^2}-(4 \sin 2 x) \frac{d y}{d x}-4(\sin 2 x \\ +\cos 2 x) y=0\end{array}$

Solution

$y=A e^x+B \sin 2 x \Rightarrow \frac{d y}{d x}=A e^x+2 B \cos 2 x$ $\frac{d^2 y}{d x^2}=A e^x-4 B \sin 2 x$ $\therefore \quad(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}=(\cos 2 x-\sin 2 x)$ $\times\left(A e^x-4 B \sin 2 x\right)$ ...(1) $\Rightarrow 4 \sin 2 x \frac{d y}{d x}=4 \sin 2 x\left(A e^x+2 B \cos x\right)$ ...(2) $\Rightarrow 4(\sin 2 x+\cos 2 x) y=4(\sin 2 x+\cos 2 x)$ $\times\left(A e^x+B \sin 2 x\right)$ ...(3) $$ \begin{aligned} & \text { Equations (1) + (2) - (3), } \\ & (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+4 \sin 2 x \frac{d y}{d x}-4(\sin 2 x+\cos 2 x) y \\ & =\left(A e^x-4 B \sin 2 x\right)(\cos 2 x-\sin 2 x)+\left(A e^x+2 B \cos 2 x\right) \\ & 4 \sin 2 x-\left(A e^x+B \sin 2 x\right)(4)(\sin 2 x+\cos 2 x) \\ & =A e^x(\cos 2 x-\sin 2 x+4 \sin 2 x-4 \sin 2 x-4 \cos 2 x) \\ & +B\left(-4 \sin 2 x \cos 2 x+4 \sin ^2 2 x+8 \sin 2 x \cos 2 x\right. \\ & \left.-4 \sin ^2 2 x-4 \sin 2 x \cos 2 x\right) \\ & =A e^x(0)+B(0)=0 . \\ & \end{aligned} $$ $\therefore$ Solution is $$ \begin{aligned} (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2} & +4 \sin 2 x \frac{d y}{d x} \\ & -4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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