If the solve the problem


If $x^{3} d y+x y d x=x^{2} d y+2 y d x ; y(2)=e$ and $x$ $>1$, then $\mathrm{y}(4)$ is equal to :

  1. $\frac{3}{2}+\sqrt{\mathrm{e}}$

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

  3. $\frac{1}{2}+\sqrt{\mathrm{e}}$

  4. $\frac{\sqrt{\mathrm{e}}}{2}$

Correct Option: , 2


$x^{3} d y+x y d x=x^{2} d y+2 y d x$

$\Rightarrow \quad \mathrm{dy}\left(\mathrm{x}^{3}-\mathrm{x}^{2}\right)=\mathrm{dx}(2 \mathrm{y}-\mathrm{xy})$

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

$\Rightarrow \quad-\ell$ ny $=\int\left(\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}^{2}}+\frac{\mathrm{C}}{(\mathrm{x}-1)}\right) \mathrm{d} \mathrm{x}$

Where $\mathrm{A}=1, \mathrm{~B}=+2, \mathrm{C}=-1$

$\Rightarrow \quad-\ell \mathrm{ny}=\ell \mathrm{n} \mathrm{x}-\frac{2}{\mathrm{x}}-\ell \mathrm{n}(\mathrm{x}-1)+\lambda$

$\Rightarrow \quad y(2)=e$

$\Rightarrow \quad-1=\ell \mathrm{n} 2-1-0+\lambda$

$\therefore \quad \lambda=-\ell \mathrm{n} 2$

$\Rightarrow \quad \ell \mathrm{n} \mathrm{y}=-\ell \mathrm{n} \mathrm{x}+\frac{2}{\mathrm{x}}+\ell \mathrm{n}(\mathrm{x}-1)+\ell \mathrm{n} 2$

Now put $x=4$ in equation

$\Rightarrow \quad \ell \mathrm{n} \mathrm{y}=-\ell \mathrm{n} 4+\frac{1}{2}+\ell \mathrm{n} 3+\ell \mathrm{n} 2$

$\Rightarrow \quad \ell \mathrm{n} \mathrm{y}=\ell \mathrm{n}\left(\frac{3}{2}\right)+\frac{1}{2} \ell \mathrm{n} \mathrm{e}$

$\Rightarrow \quad y=\frac{3}{2} \sqrt{e}$

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