Solve (x+2y+7)y'+2x-y+4=0

Solve $(x+2y+7)y'+2x-y+4=0$.

Step 1: Confirm the intersection point

\texttt{sp.solve}$([x+2y+7,\ 2x-y+4],[x,y])$ returns $\{x=-3,\ y=-2\}$, matching the source.

Step 2: Confirm the homogeneous reduction

With $x=X-3,\ y=Y-2$, \texttt{sympy} substitutes into $\dfrac{dy}{dx}=-\dfrac{2x-y+4}{x+2y+7}$ and simplifies to $\dfrac{dY}{dX}=-\dfrac{2X-Y}{X+2Y}$, matching the source. Setting $Y=vX$ and simplifying with \texttt{sympy} confirms \[X\frac{dv}{dX}=\frac{-2(1+v^2)}{1+2v}\ \Longrightarrow\ \frac{1+2v}{1+v^2}\,dv=-\frac{2\,dX}{X},\] exactly as claimed.

Step 3: Confirm the integration

\texttt{sp.integrate}$\!\left(\dfrac{1+2v}{1+v^2},v\right)=\arctan v+\ln(1+v^2)$, matching the source term-by-term, giving $\arctan v+\ln[(1+v^2)X^2]=C_1$, i.e. $\arctan\dfrac{Y}{X}+\ln(X^2+Y^2)=C_1$ after back-substituting $v=Y/X$.

Step 4: Supply the missing final verification (implicit differentiation)

The source never checks its boxed answer against the original ODE — unlike Problems 213/215, no verification step is given. Implicit-differentiating \[F(x,y)=\arctan\frac{y+2}{x+3}+\ln\left[(x+3)^2+(y+2)^2\right]-C=0\] with \texttt{sympy} (solving $F_x+F_y\,y'=0$ for $y'$) gives, after simplification, \[y'=\frac{-2x+y-4}{x+2y+7}=-\frac{2x-y+4}{x+2y+7},\] which \texttt{sympy} confirms is identical to the original equation's own $y'=-\dfrac{2x-y+4}{x+2y+7}$, i.e. $(x+2y+7)y'+2x-y+4=0$ exactly — residual $0$. The source's unverified answer is confirmed genuinely correct.