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Question

# Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.

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Solution

## According to the question, $\frac{dy}{dx}=x+xy\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=x\left(1+y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{y+1}dy=xdx\phantom{\rule{0ex}{0ex}}\mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\int \frac{1}{y+1}dy=\int xdx\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}\left|y+1\right|=\frac{{x}^{2}}{2}+\mathrm{log}C\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}\left|\frac{y+1}{C}\right|=\frac{{x}^{2}}{2}\phantom{\rule{0ex}{0ex}}⇒y+1=C{e}^{\frac{{x}^{2}}{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Since},\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(0,1\right).\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}.\phantom{\rule{0ex}{0ex}}\therefore 1+1=C{e}^{0}\phantom{\rule{0ex}{0ex}}⇒C=2\phantom{\rule{0ex}{0ex}}\mathrm{Puting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}.\mathrm{We}\mathrm{get}\phantom{\rule{0ex}{0ex}}y+1=2{e}^{\frac{{x}^{2}}{2}}\phantom{\rule{0ex}{0ex}}⇒y=-1+2{e}^{\frac{{x}^{2}}{2}}$

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