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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 the ordinate of the point.

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Toolbox:
• Slope of the tangent to the curve is $\large\frac{dy}{dx}$
• If the linear differential equation is of the form $\large \frac{dy}{dx}$$=f(x) then it can be solved by seperating the variables and then integrating. • \large \int \frac{dx}{a^2-x^2}=\frac{1}{2a}$$\log \large\frac{|a+x|}{|a-x|}$
Step 1:
Given :The square of the difference of the abscissa and the ordinate is equal to the slope of the tangent of the curve.
Slope of the tangent to the curve is $\large\frac{dy}{dx}$
The square of the difference of the abscissa and the ordinate is $(x-y)^2$
Hence $\large\frac{dy}{dx}$$=(x-y)^2 Put x-y=z on differentiating both sides w.r.t x we get, 1-\large\frac{dy}{dx}=\large\frac{dz}{dx}=>\frac{dy}{dx}$$=1-\large\frac{dz}{dx}$
Now substituting this we get,
$1-\large\frac{dz}{dx}$$=z^2 rearranging this we get, \large\frac{dz}{dx}$$=1-z^2$
Now seperating the variables we get,
$\large\frac{dz}{1-z^2}$$=dx Step 2: Integrating on both sides, \large \int \frac{dz}{1-z^2}$$=\int dx$
$\large \int \frac{dz}{1-z^2}$ is of the form $\large \int \frac{dx}{a^2-x^2}=\frac{1}{2a} $$\log \large\frac{|a+x|}{|a-x|}$$+c$
Here $a=1$ and $x=z$
$\int \large\frac{dz}{1-z^2}=\frac{1}{2} $$\log \bigg|\large\frac{1+z}{1-z}\bigg| Therefore \large \frac{1}{2}$$\log \bigg|\large\frac{1+z}{1-z}\bigg|$$=x+c \log_e \bigg|\large\frac{1+z}{1-z}\bigg|$$=2x+c$
Substituting for $z=x-y$
$\log_e \bigg|\large\frac{1+(x-y)}{1-x+y}\bigg|$$=2x+c$
$ce^{2x}=\large\frac{1+(x-y)}{(1-x+y)}$
$ce^{2x}(1-x+y)=(1+(x-y)$
It is given the curve passes through the orgin. Hence substitute $x=0$ and $y=0$ to evaluate the value of c
$ce^0(1-0+0)=1+(0-0)$
But $e^0=1$
$c=1$
Therefore $e^{2x}(1-x+y)=1+x-y$
$e^{2x}(1-x+y)-1=x-y$
answered May 24, 2013 by