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If x^{3}dy+xy\; dx=x^{2}dy+2y\; dx;\; y(2)=e and x> 1, then y(4) is equal to :
Option: 1 \frac{3}{2}+\sqrt{e}  
Option: 2 \frac{3}{2}\sqrt{e}
Option: 3 \frac{1}{2}+\sqrt{e}
Option: 4 \frac{\sqrt{e}}{2}

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\\ x^{3} d y+x y d x=x^{2} d y+2 y d x \\ \Rightarrow d y\left(x^{3}-x^{2}\right)=d x(2 y-x y) \\ \Rightarrow-\int \frac{1}{y} d x=\int \frac{x-2}{x^{2}(x-1)} d x \\ \Rightarrow-\ell n y=\int\left(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-1)}\right) d x

Where A = 1, B = + 2, C = –1

\begin{array}{l} \Rightarrow-\ell n y=\ell n x-\frac{2}{x}-\ln (x-1)+\lambda \\ \Rightarrow y(2)=e \\ \Rightarrow-1=\ln 2-1-0+\lambda \\ \therefore \lambda=-\ln 2 \end{array}

\begin{aligned} &\Rightarrow \ell \mathrm{ny}=-\ell \mathrm{nx}+\frac{2}{\mathrm{x}}+\ln (\mathrm{x}-1)+\ln 2\\ &\text { Now put } x=4 \text { in equation }\\ &\Rightarrow \ell \mathrm{ny}=-\ln 4+\frac{1}{2}+\ln 3+\ln 2\\ &\Rightarrow \quad \ell \mathrm{ny}=\ln \left(\frac{3}{2}\right)+\frac{1}{2} \ell \mathrm{ne}\\ &\Rightarrow \mathrm{y}=\frac{3}{2} \sqrt{\mathrm{e}} \end{aligned}

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himanshu.meshram

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