Separable Equations — Question 10

Question 10 – Solution

Strategy. Use the zero terminal slope to constrain the coefficient, then use the separated integral to enforce the target height.

Step 1: Determine the coefficient. Since $1-y(1)^2=3/4$, $y'(1)=0$ forces $\alpha+\beta=0$. The smooth right-hand side gives local uniqueness, so the solution starting at $0$ cannot meet either constant solution $y=\pm 1$. On its branch $-1<y<1$, $$\frac 12\ln\frac{1+y}{1-y}=\alpha x+\frac{\beta x^2}{2}.$$ At $x=1$, this becomes $\tfrac 12\ln 3=\alpha+\beta/2$. Hence $$\boxed{\alpha=\ln 3,\qquad\beta=-\ln 3}.$$ These two independent linear conditions have exactly one pair of coefficients.

Step 2: Solve and interpret the curve. Set $E(x)=3^{\,2x-x^2}>0$. Exponentiating and isolating $y$ gives $$\boxed{y(x)=\frac{E(x)-1}{E(x)+1},\qquad I=\mathbb R}.$$ Since $E+1>0$, the formula is globally smooth and lies strictly between $-1$ and $1$. Its zeros are $x=0,2$. The exponent $2x-x^2=1-(x-1)^2$ has its unique maximum at $1$, and $(E-1)/(E+1)$ increases with $E$. Thus the unique global maximum is $y(1)=1/2$, while both limits at infinity are $-1$.

Step 3: Verify and restore constants. Here $E'=2(\ln 3)(1-x)E$ and $1-y^2=4E/(E+1)^2$, so $y'=(\ln 3)(1-x)(1-y^2)$. Also $E(0)=1$ and $E(1)=3$ verify both values and the zero slope. The omitted constants are $\boxed{y\equiv 1,\ y\equiv-1}$; neither satisfies $y(0)=0$.