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  • Let the solution curve of the differential equation <img alt="\mathrm{x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0}," src="/latex-image/?%5Cmathrm%7Bx%20%5Cmathrm%7B%

Let the solution curve of the differential equation \mathrm{x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0}, intersect the line \mathrm{x=1 \text { at } y=0} at  \mathrm{y=0 } and the line  \mathrm{x=2 \text { at } y=\alpha}. Then the value of \mathrm{\alpha } is:

Option: 1

\frac{1}{2}


Option: 2

\frac{3}{2}


Option: 3

-\frac{3}{2}


Option: 4

\frac{5}{2}


Answers (1)

best_answer

\begin{aligned} & \mathrm{x \;dy=\left(\sqrt{x^{2}+y^{2}}+y\right) dx} \\ & \mathrm{x \; dy-y \;dx=\sqrt{x^{2}+y^{2}} d x} \\ &\mathrm{\frac{x\; d y-y\; d x}{x^{2}}=\sqrt{1+\frac{y^{2}}{x^{2}}} \cdot \frac{d x}{x} }\\ & \mathrm{ \frac{d\left(\frac{y}{x}\right)}{\sqrt{1+\left(\frac{y}{x}\right)^{2}}}=\frac{d x}{x}} \\ & \mathrm{ \ln \left(\frac{y}{x}+\sqrt{\left.\left(\frac{y}{x}\right)^{2}+1\right)}=\ln x+\ln c\right.} \\ & \mathrm{ \frac{y+\sqrt{y^{2}+x^{2}}}{x}=c x} \\ &\mathrm{y+\sqrt{y^{2}+x^{2}}=c x^{2}} \\ & \mathrm{x=1, y=0 \Rightarrow 0+1=c \Rightarrow c=1} \\ & \mathrm{\text { curve is } y+\sqrt{x^{2}+y^{2}}=x^{2}} \\ & \mathrm{x=2, y=\alpha} \\ & \mathrm{2+\sqrt{4+\alpha^{2}}=4} \\ & \mathrm{4+\alpha^{2}=16+\alpha^{2}=8 \alpha} \\ & \mathrm{\alpha=\frac{3}{2}} \end{aligned}

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himanshu.meshram

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