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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
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The correct answer is:
$\begin{aligned} &(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x}-4(\sin 2 x \\ &+\cos 2 x) y=0\end{aligned}$
$y=A e^x+B \sin 2 x \Rightarrow \frac{d y}{d x}=A e^x+2 B \cos 2 x$
$\frac{d^2 y}{d x^2}=A e^x-4 B \sin 2 x$
$\therefore \quad(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}=(\cos 2 x-\sin 2 x)$
$\times\left(A e^x-4 B \sin 2 x\right)$ ...(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
$$
\begin{aligned}
(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2} & +4 \sin 2 x \frac{d y}{d x} \\
& -4(\sin 2 x+\cos 2 x) y=0
\end{aligned}
$$
$\frac{d^2 y}{d x^2}=A e^x-4 B \sin 2 x$
$\therefore \quad(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}=(\cos 2 x-\sin 2 x)$
$\times\left(A e^x-4 B \sin 2 x\right)$ ...(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
$$
\begin{aligned}
(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2} & +4 \sin 2 x \frac{d y}{d x} \\
& -4(\sin 2 x+\cos 2 x) y=0
\end{aligned}
$$
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