Differential Equation - how to solve for the equation of a curve passing through points? Find the equation of a curve passing through. $(1,1)$ where the $x$-intercept of the tangent line from point $(x,y)$ is equal to the $y$-intercept of the normal line from the point.
is this correct?
$$\frac{dy}{dx}=xy$$ 
is the slope xy?
 A: The ODE you obtained is wrong. Write the curve $\gamma$ in question in terms of the polar angle $\phi$ as follows:
$$\gamma:\quad \phi\mapsto\left\{\eqalign{u(\phi)&=r(\phi)\cos\phi\cr v(\phi)&=r(\phi)\sin\phi\cr}\right.\ .$$
For given parameter value $\phi$ the tangent to $\gamma$ is then given by
$$t\mapsto(u+t\dot u,v+t\dot v)\qquad(-\infty<t<\infty)$$
and intersects the $x$-axis at $\xi=u-{\dot u v\over \dot v}$. Similarly, the normal to $\gamma$ is given by
$$t\mapsto(u-t\dot v,v+t\dot u)\qquad(-\infty<t<\infty)$$
and intersects the $y$-axis at $\eta=v+{u\dot u\over\dot v}$. The condition $\xi=\eta$ then leads to
$$\dot u(u+v)=\dot v(u-v)\ .\tag{1}$$
Now
$$\dot u=\dot r\cos\phi-r\sin\phi,\quad\dot v=\dot r\sin\phi+r\cos\phi\ .$$
When we plug this into $(1)$ a factor $r$ drops out, and we obtain
$$(\dot r\cos\phi-r\sin\phi)(\cos\phi+\sin\phi)=(\dot r\sin\phi+r\cos\phi)(\cos\phi-\sin\phi)\ .$$
Now a miracle happens: This simplifies to
$$\dot r=r$$
and implies that $\gamma$ is an arc of a logarithmic spiral: $r(\phi)=Ce^\phi$. The condition $r\bigl({\pi\over4}\bigr)=\sqrt{2}$ then determines the constant $C$, and we obtain definitively
$$r(\phi)=\sqrt{2}\>e^{\phi-\pi/4}\qquad\left(0<\phi<{\pi\over2}\right)\ .$$
A: Remember that you can re-write your DE to
$$\frac{dy}{y}=xdx$$
Which can be readily solved with elementary functions. Please let me know if you cannot and I'll amend the hint.
On the second aprt of the question; once you have a general solution incorporating a constant, say, $C$ the line passing through $(1,1)$ will allow you to find a particular solution to the ODE.
Good luck!
