Let y=y(x) be the solution of the differential equation [(dy)/(dx) = (1+x+x^2)(1-y+y^2), y…

Question

Let $y=y(x)$ be the solution of the differential equation
$$\frac{dy}{dx} = (1+x+x^2)(1-y+y^2),\qquad y(0)=\frac12.$$
Then $2y(1)-1$ is equal to:

Accepted Answer

Separating variables:
$$\int\frac{dy}{1-y+y^2} = \int(1+x+x^2)\,dx.$$

Since $1-y+y^2 = \left(y-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2$,
$$\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y-1}{\sqrt3}\right) = x+\frac{x^2}{2}+\frac{x^3}{3}+C.$$

Using $y(0)=1/2$: $C=0$.

At $x=1$: $x+\frac{x^2}{2}+\frac{x^3}{3}=\frac{11}{6}$, so
$$\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y(1)-1}{\sqrt3}\right)=\frac{11}{6} \implies \tan^{-1}\!\left(\frac{2y(1)-1}{\sqrt3}\right) = \frac{11\sqrt3}{12}.$$

So $2y(1)-1 = \sqrt3\tan\!\left(\dfrac{11\sqrt3}{12}\right)$.

Solutions · 1