If x^{3} dy + xy dx = x^{2} dy + 2y dx,y(2) = | vedclass

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(B) Given equation: $x^{3} dy + xy dx = x^{2} dy + 2y dx$Rearranging terms: $(x^{3} - x^{2}) dy = (2y - xy) dx$$(x^{3} - x^{2}) dy = y(2 - x) dx$Separ

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$\frac{3}{2} \sqrt{e}$

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(B) Given equation: $x^{3} dy + xy dx = x^{2} dy + 2y dx$Rearranging terms: $(x^{3} - x^{2}) dy = (2y - xy) dx$$(x^{3} - x^{2}) dy = y(2 - x) dx$Separating variables: $\frac{dy}{y} = \frac{2 - x}{x^{2}(x - 1)} dx$Using partial fractions: $\frac{2 - x}{x^{2}(x - 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 1}$$2 - x = Ax(x - 1) + B(x - 1) + Cx^{2}$For $x = 0$,$2 = -B \Rightarrow B = -2$. For $x = 1$,$1 = C$. Comparing $x^{2}$ coefficients,$0 = A + C \Rightarrow A = -1$.Integrating: $\int \frac{dy}{y} = \int \left( -\frac{1}{x} - \frac{2}{x^{2}} + \frac{1}{x - 1} \right) dx$$\ln y = -\ln x + \frac{2}{x} + \ln(x - 1) + C_{1}$Given $y(2) = e$: $\ln e = -\ln 2 + \frac{2}{2} + \ln(2 - 1) + C_{1} \Rightarrow 1 = -\ln 2 + 1 + 0 + C_{1} \Rightarrow C_{1} = \ln 2$.So,$\ln y = \ln \left( \frac{2(x - 1)}{x} \right) + \frac{2}{x}$.For $x = 4$: $\ln y = \ln \left( \frac{2(3)}{4} \right) + \frac{2}{4} = \ln \left( \frac{3}{2} \right) + \frac{1}{2} = \ln \left( \frac{3}{2} \right) + \ln \sqrt{e}$.$y = \frac{3}{2} \sqrt{e}$.

If $x^{3} dy + xy dx = x^{2} dy + 2y dx$,$y(2) = e$ and $x > 1$,then $y(4)$ is equal to

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