Find the general solution of y'=(2xy+1)/(y-x^2) , and…
Find the general solution of $y'=\dfrac{2xy+1}{y-x^2}$, and determine the integral curve(s) that intersect the line $y=-2$ at an angle of $45^\circ$.
Explanation of Method
Rewrite as $M\,dx+N\,dy=0$, verify exactness, integrate to find the potential function, then impose the slope condition $y'=\tan45^\circ=1$ at $y=-2$ to pin down the constant.
Step 1: Rewrite and check exactness
$(2xy+1)\,dx+(x^2-y)\,dy=0$, with $M=2xy+1$, $N=x^2-y$: $M_y=2x=N_x$, so the equation is exact.
Step 2: Integrate to find the potential function
$F=\displaystyle\int(2xy+1)\,dx=x^2y+x+g(y)$; matching $F_y=x^2+g'(y)$ to $N=x^2-y$ gives $g'(y)=-y$, so $g(y)=-\dfrac{y^2}{2}$.
Step 3: General solution
\[\boxed{\,2x^2y+2x-y^2=C\,}\]
Step 4: Apply the $45^\circ$ angle condition
At $y=-2$ the slope must be $\tan45^\circ=1$: $\dfrac{2x(-2)+1}{-2-x^2}=1\ \Rightarrow\ -4x+1=-2-x^2\ \Rightarrow\ x^2-4x+3=0\ \Rightarrow\ (x-1)(x-3)=0$, so $x=1$ or $x=3$.
Step 5: Determine the constants
At $(1,-2)$: $C=2(1)(-2)+2(1)-4=-6$. At $(3,-2)$: $C=2(9)(-2)+2(3)-4=-34$.
Result
Final answer:
General solution: \[\boxed{\,2x^2y+2x-y^2=C\,}\]
Curves meeting $y=-2$ at $45^\circ$: \[\boxed{\,2x^2y+2x-y^2=-6\ \text{(at }(1,-2)\text{)}\qquad\text{and}\qquad 2x^2y+2x-y^2=-34\ \text{(at }(3,-2)\text{)}\,}\]
(verified with sympy: $F_x=M$, $F_y=N$ identically, and both slopes evaluate to exactly $1$.)