For all x 0,let y_1(x), y_2(x),and y_3(x) b | vedclass

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(A) Given the differential equations:$1) \frac{dy_1}{y_1} = \sin^2 x dx \implies \ln y_1 = \int \sin^2 x dx + C_1$$2) \frac{dy_2}{y_2} = \cos^2 x dx \

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$2$

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(A) Given the differential equations:$1) \frac{dy_1}{y_1} = \sin^2 x dx \implies \ln y_1 = \int \sin^2 x dx + C_1$$2) \frac{dy_2}{y_2} = \cos^2 x dx \implies \ln y_2 = \int \cos^2 x dx + C_2$$3) \frac{dy_3}{y_3} = \left(\frac{2}{x^3} - 1\right) dx \implies \ln y_3 = \int (2x^{-3} - 1) dx + C_3 = -x^{-2} - x + C_3$Summing the equations: $\ln(y_1 y_2 y_3) = \int (\sin^2 x + \cos^2 x + \frac{2}{x^3} - 1) dx + C = \int (1 + \frac{2}{x^3} - 1) dx + C = \int \frac{2}{x^3} dx + C = -x^{-2} + C$.Using initial conditions at $x=1$: $\ln(5 \cdot \frac{1}{3} \cdot \frac{3}{5e}) = \ln(e^{-1}) = -1$.So,$-1 = -(1)^{-2} + C \implies C = 0$.Thus,$y_1 y_2 y_3 = e^{-1/x^2}$.Now,evaluate $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2} + 2x}{e^{3x} \sin x} = \lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{e^{3x} \sin x} + \lim_{x \rightarrow 0^+} \frac{2x}{e^{3x} \sin x}$.Since $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{\sin x} = 0$ and $\lim_{x \rightarrow 0^+} \frac{2x}{\sin x} = 2$,the limit is $0 + 2 = 2$.

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