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.
Abscissa means the x-coordinate and ordinate means the y-coordinate
Slope of tangent is given as square of the difference of the abscissa and ordinate
Difference of abscissa and ordinate is (x – y) and its square will be (x – y)2
Hence slope of tangent is (x – y)2
Slope of tangent of a curve y = f(x) is given by
Substitute x – y = z hence y = x – z
Differentiate x – z with respect to x
Using partial fraction for
Equating numerator
⇒ A(1 – z) + B(1 + z) = 1
Put z = 1
⇒ B = 1/2
Put z = -1
⇒ A = 1/2
Hence
Put in (a)
Integrate
⇒ x = 1/2(log(1 + z) + (–log(1 – z)) + c
⇒ 2x = log(1 + z) – log(1 – z) + c
Using log a – log b = log
Resubstitute z = x – y
Now it is given that the curve is passing through origin that is (0, 0)
Hence (0, 0) will satisfy the curve equation (b)
Putting values x = 0 and y = 0 in (b)
⇒ c = 0
Put c = 0 back in equation (b)
⇒ e2x(1 – x + y) = (1 + x – y)
Hence equation of curve is e2x(1 – x + y) = (1 + x – y)