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(A) Given $y=A e^x+B \sin 2 x$.Differentiating with respect to $x$:$\frac{d y}{d x}=A e^x+2 B \cos 2 x$ ...$(1)$Differentiating again:$\frac{d^2 y}{d

Vedclass · Accepted Answer

$\begin{aligned} &(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x}-4(\sin 2 x+\cos 2 x) y=0\end{aligned}$

Vedclass · Answer

(A) Given $y=A e^x+B \sin 2 x$.Differentiating with respect to $x$:$\frac{d y}{d x}=A e^x+2 B \cos 2 x$ ...$(1)$Differentiating again:$\frac{d^2 y}{d x^2}=A e^x-4 B \sin 2 x$ ...$(2)$From $(1)$, $A e^x = \frac{d y}{d x} - 2 B \cos 2 x$. Substituting this into $(2)$:$\frac{d^2 y}{d x^2} = \frac{d y}{d x} - 2 B \cos 2 x - 4 B \sin 2 x$.By testing the options or eliminating constants $A$ and $B$, we find that the differential equation is $(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+4 \sin 2 x \frac{d y}{d x}-4(\sin 2 x+\cos 2 x) y=0$.

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

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