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  • Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

Answers (1)

Abscissa refers to the x coordinate and ordinate refers to the y coordinate.

 Slope of the tangent is the square of the difference of the abscissa and the ordinate.

Difference of the abscissa and ordinate is (x-y) and its square is (x-y) ^2

Hence the Slope of the tangent is (x-y) ^2

\frac{d y}{d x}=(x-y)^{2}

\begin{aligned} &\text { Put } \quad x-y=z\\ &\Rightarrow \quad 1-\frac{d y}{d x}=\frac{d z}{d x}\\ &\Rightarrow \quad 1-\frac{d z}{d x}=\frac{dy}{dx}=z^{2}\\ &\Rightarrow \quad 1-z^{2}=\frac{d z}{d x}\\ &\Rightarrow \quad d x=\frac{d z}{1-z^{2}}\\ &\Rightarrow \quad \int d x=\int \frac{d z}{1-z^{2}} \end{aligned}

\\x=\frac{1}{2} \log \left|\frac{1+z}{1-z}\right|+C \\ x=\frac{1}{2} \log \left| \frac{1+x-y}{1-x+y}\right|+C

The curve passes through the (0,0)

\\ 0=\frac{1}{2} \log 1+C \\ C=0
\\ \Rightarrow \mathrm{e}^{2 \mathrm{x}}=\frac{1+\mathrm{x}-\mathrm{y}}{1-\mathrm{x}+\mathrm{y}} \\ \Rightarrow \mathrm{e}^{2 x}(1-\mathrm{x}+\mathrm{y})=(1+\mathrm{x}-\mathrm{y})

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