Question

Find the equation of a curve in the xy-plane that passes through (1,3) and whose tangent at a point (x,y) has a slope (-6x(2+y))/√(〖3-3x〗^2 )

Answer 1

Slope at (x,y)

$\dfrac{dy}{dx}=\dfrac{-6x(2+y)}{(3-3x)^2}$

Solving above differential equation:

It is first order linear ordinary differential equation

$\dfrac{dy}{(2+y)}=\dfrac{-6x}{(3-3x)^2}\\\ \\\int \dfrac{dy}{(2+y)}=\int \dfrac{-6x}{(3-3x)^2}\\\ \\ln|y+2|=2\sqrt{3-3x^2}\ +C$

and the curve passes through (1,3)

$\ln|y+2|=2\sqrt{3-3x^2}\ +C\\\ln|3+2|=2\sqrt{3-3(1)^2}+c\\\ln|5|=C$

Hence, C = ln|5|

So putting this value in above equation, we get

$\ln|y+2|=2\sqrt{3-3x^2}+ln5\\y+2=e^{2\sqrt{3-3x^2}+ln5}\\y=e^{2\sqrt{3-3x^2}+ln5}-2$

Hence equation of curve = $\boxed{y=e^{2\sqrt{3-3x^2}+ln5}-2}$