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Question

# At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.

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Solution

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

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