Answer to Question #268218 in Differential Equations for jay

Question

1. Find the equation of the curve at every point that passes through the point (0,-1) and which the normal line at any point (x,y) has a slope of – 1/2(x+1). Ans. y=x²+2x-1
2. Given the family of curves x²+y²=cx, find the family of curves of the orthogonal trajectories. Ans. x²+y²=ky

Answer 1

1.

$$slope_1=m_1=-\dfrac{1}{2x+1}$$

Then the tangent line at any point$(x,y)$ has a slope  

$$slope_2=m_2=-\dfrac{1}{m_1}=2x+1$$

$$y'=2x+1$$

Integrate

$$y=\int(2x+1)dx=x^2+x+C$$

The curve passes through the point $(0,-1)$

$$-1=(0)^2+0+C=>C=-1$$

The equation of the curve is

$$y=x^2+x-1$$

2.  Given family of curves is $x^2+y^2=cx$

Differentiate both sides with respect to $x$

$$2x+2yy'=c$$

$$y'=\dfrac{c-2x}{2y}$$

$$y'=\dfrac{x^2+y^2-2x^2}{2xy}$$

$$y'=\dfrac{y^2-x^2}{2xy}$$

If any curve intersects orthogonally at point $(x,y),$ then (if its slope is $y'$) we must have

$$\dfrac{dx}{dy}=-\dfrac{y^2-x^2}{2xy}=\dfrac{x}{2y}-\dfrac{y}{2x}$$

Solving the above differential equation, we get

$$x=ty$$

$$\dfrac{dx}{dy}=t+y\dfrac{dt}{dy}$$

$$t+y\dfrac{dt}{dy}=\dfrac{t}{2}-\dfrac{1}{2t}$$

$$y\dfrac{dt}{dy}=-\dfrac{t}{2}-\dfrac{1}{2t}$$

$$t\dfrac{dt}{t^2+1}=-\dfrac{dy}{2y}$$

Integrate

$$\int t\dfrac{dt}{t^2+1}=-\int\dfrac{dy}{2y}$$

$$t^2+1=\dfrac{k}{y}$$

$$(\dfrac{x}{y})^2+1=\dfrac{k}{y}$$

$$x^2+y^2=ky$$