if $x^{3} d y + x y d x = x^{2} d y + 2 y d x; y (2) = e and x > 1, then y (4) =$

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

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

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

$\frac{1}{2} + \sqrt{e}$

answer is B.

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Detailed Solution

$x^{3} d y + x y d x = 2 y d x + x^{2} d y \Rightarrow (x^{3} - x^{2}) d y = (2 y - x y) d x \Rightarrow \frac{d y}{y} = \frac{2 - x}{x^{2} (x - 1)} d x$

let $\frac{2 - x}{x^{2} (x - 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 1} . Then A x (x - 1) + B (x - 1) + C x^{2} = 2 - x$

$\begin{array}{l} x = 1 \Rightarrow C = 1; x = 0 \Rightarrow - B = 2 \Rightarrow B = - 2 \\ x = 2 \Rightarrow 2 A + B + 4 C = 0 \Rightarrow 2 A - 2 + 4 = 0 \Rightarrow A = - 1 \end{array}$
$\begin{array}{l} ∴ \frac{d y}{y} = \frac{2 - x}{x^{2} (x - 1)} d x \Rightarrow \int \frac{d y}{y} = \int (- \frac{1}{x} - \frac{2}{x^{2}} + \frac{1}{x - 1}) d x \Rightarrow \log y = - \log x + \frac{2}{x} + \log (x - 1) + \log c \\ y (2) = e \Rightarrow \log e = - \log 2 + 1 + \log 1 + \log c \Rightarrow c = 2 . \\ ∴ \log y = - \log x + \frac{2}{x} + \log (x - 1) + \log 2 \\ If x = 4 then \log y (4) = - \log 4 + \frac{2}{4} + \log 3 + \log 2 \Rightarrow y (4) = \frac{3}{2} e^{1 / 2} . \end{array}$