Problem 1.
A force field given by
\[
\mathbf{F}(x,y)=\frac{2y}{\sqrt{x^2+y^2}}\mathbf{i}-\frac{y^2-x}{\sqrt{x^2+y^2}}\mathbf{j}.
\]
Draw it by finding and sketching the family of curves tangent to $\mathbf{F}$.
At the point $(x,y)$ in the plane, the vector $\mathbf{F}(x,y)$ has a slope of
\[
\frac{dy}{dx}=\frac{-(y^2-x)/\sqrt{x^2+y^2}}{2y/\sqrt{x^2+y^2}}=\frac{-(y^2-x)}{2y}
\]
which, in differential form, is
\begin{align*}
2y\,dy &=-(y^2-x),\,dx\\
(y^2-x)\,dx+2y\,dy&=0.
\end{align*}
The given equation is not exact because $M_y(x,y)=2y$ and $N_x(x,y)=0$. However, because
\[
\frac{M_y(x,y)-N_x(x,y)}{N(x,y)}=\frac{2y-0}{2y}=1=h(x)
\]
it follows that $e^{\int h(x)\,dx}=e^{\int 1\,dx}=e^x$ is an integrating factor. Multiplying the given
differential equation by $e^x$ produces the exact differential equation
\[
(y^2e^x-xe^x)\,dx+2ye^x\,dy=0
\]
whose solution is obtained as follows.
After integration we obtain $g(x)=-xe^x+e^x+C_1$, which implies that \[ f(x,y)=y^2e^x-xe^x+e^x+C_1. \] The general solution is $y^2e^x-xe^x+e^x=C$, or $y^2=x-1+Ce^{-x}$. The force vector at $(x,y)$ is tangent to the curve passing through $(x,y)$.
\begin{align*}
&\Step{1A}{f(x,y)=\int N(x,y)\,dy=\int 2ye^x\,dy=y^2e^x+g(x),}\\
&\Step{2A}{f_x(x,y)=y^2e^x+g'(x),} \\
&\Step{3A}{f_x(x,y)=M(x,y)=y^2e^x-xe^x,} \\
&\Step{4A}{g'(x)=-xe^x.}\\
\end{align*}
After integration we obtain $g(x)=-xe^x+e^x+C_1$, which implies that \[ f(x,y)=y^2e^x-xe^x+e^x+C_1. \] The general solution is $y^2e^x-xe^x+e^x=C$, or $y^2=x-1+Ce^{-x}$. The force vector at $(x,y)$ is tangent to the curve passing through $(x,y)$.
