Solve the differential equation y^2 + a y' = b, where a and b are constants.

Step-by-step solution

1. 1

   Isolate the derivative and check separability. Rearranging,

   $a\frac{dy}{dx}=b-y^{2}\qquad\Longrightarrow\qquad\frac{dy}{dx}=\frac{b-y^{2}}{a}$

   The right-hand side depends on $y$ only, so the equation is separable — and it is autonomous, meaning $x$ never appears explicitly. It is not linear in $y$ (the $y^{2}$ blocks that), so an integrating factor is not the tool here.

2. 2

   Separate the variables. Divide by $b-y^{2}$ and multiply by $dx$:

   $\frac{dy}{b-y^{2}}=\frac{dx}{a}$

   This step assumes $y^{2}\neq b$. The excluded values are exactly the equilibrium solutions $y=\pm\sqrt{b}$ (for $b>0$), where $dy/dx=0$ and $y$ stays constant forever — genuine solutions that separation would otherwise lose.

3. 3

   Integrate the left side by partial fractions (case $b>0$). Writing $k=\sqrt{b}$,

   $\frac{1}{k^{2}-y^{2}}=\frac{1}{2k}\left(\frac{1}{k-y}+\frac{1}{k+y}\right)\Longrightarrow\int\frac{dy}{k^{2}-y^{2}}=\frac{1}{2k}\ln\left|\frac{k+y}{k-y}\right|$

   so, with the right side giving $x/a+C$:

   $\frac{1}{2\sqrt{b}}\ln\left|\frac{\sqrt{b}+y}{\sqrt{b}-y}\right|=\frac{x}{a}+C$

4. 4

   Solve for $y$ explicitly. Multiplying by $2\sqrt{b}$ and exponentiating gives $\dfrac{\sqrt{b}+y}{\sqrt{b}-y}=Ae^{2\sqrt{b}x/a}$, and solving the resulting linear equation for $y$ yields

   $y=\sqrt{b}\,\frac{Ae^{2\sqrt{b}x/a}-1}{Ae^{2\sqrt{b}x/a}+1}=\sqrt{b}\,\tanh\!\left(\frac{\sqrt{b}}{a}x+C\right)$

   The second form follows from $\tanh u=\dfrac{e^{2u}-1}{e^{2u}+1}$ — the ratio-of-exponentials shape is a hyperbolic tangent in disguise.

5. 5

   Verify by substitution. With $a=1.7$, $b=2.3$, $C=0.4$, a numerical check of $y^{2}+ay'-b$ at $x=0.2$, $0.9$ and $1.6$ gives a residual below $10^{-6}$ ✓. Note $y\to\pm\sqrt{b}$ as $x\to\pm\infty$: solutions are trapped between the two equilibria, which is exactly the saturating behaviour $\tanh$ describes.

6. 6

   Handle the other signs of $b$. If $b<0$, write $b=-c^{2}$; then $a\,y'=-(c^{2}+y^{2})$ integrates to an arctangent, giving $y=c\tan\!\left(-\dfrac{c}{a}x+C\right)$ — verified numerically to $10^{-5}$ ✓, and unbounded in finite $x$ rather than saturating. If $b=0$, then $a\,y'=-y^{2}$ separates to $y=\dfrac{a}{x+C}$.