Solve the differential equation 3(1 + t^2) dy/dt = 2ty(y^3 - 1).

Step-by-step solution

1. 1

   Spot the equilibrium solutions first. The right side vanishes when $y=0$ or $y^3=1$, and a constant $y$ has $\tfrac{dy}{dt}=0$, so

   $y\equiv 0\qquad\text{and}\qquad y\equiv 1$

   both solve the equation. Finding these up front matters, because the substitution below divides by $y$.

2. 2

   Recognise the Bernoulli shape. Dividing by $3(1+t^2)$ and expanding the right side:

   $\frac{dy}{dt}-\frac{2t}{3(1+t^2)}\,y=\frac{2t}{3(1+t^2)}\,y^{4}$

   This is Bernoulli with exponent $n=4$, and the standard cure is $v=y^{1-n}=y^{-3}$.

3. 3

   Substitute $v=y^{-3}$. Then $\dfrac{dv}{dt}=-3y^{-4}\dfrac{dy}{dt}$, and using $\dfrac{dy}{dt}=\dfrac{2ty(y^3-1)}{3(1+t^2)}$:

   $\frac{dv}{dt}=-3y^{-4}\cdot\frac{2ty(y^3-1)}{3(1+t^2)}=-\frac{2t\left(1-y^{-3}\right)}{1+t^2}$

   so

   $\frac{dv}{dt}-\frac{2t}{1+t^2}\,v=-\frac{2t}{1+t^2}$

   Note the sign: the coefficient of $v$ is negative. Getting it positive here is the single most common slip in this problem, and it flips the final answer.

4. 4

   Find the integrating factor. With $P(t)=-\dfrac{2t}{1+t^2}$,

   $\mu(t)=\exp\!\left(\int P\,dt\right)=\exp\left(-\ln\left(1+t^2\right)\right)=\frac{1}{1+t^2}$

   Multiplying through turns the left side into a single derivative:

   $\frac{d}{dt}\left(\frac{v}{1+t^2}\right)=-\frac{2t}{\left(1+t^2\right)^2}$

5. 5

   Integrate both sides. The right side is itself a substitution ($u=1+t^2$):

   $\int-\frac{2t}{\left(1+t^2\right)^2}\,dt=\frac{1}{1+t^2}+C$

   so $\dfrac{v}{1+t^2}=\dfrac{1}{1+t^2}+C$, that is

   $v=1+C\left(1+t^2\right)$

6. 6

   Return to $y$ and verify. Since $v=y^{-3}$,

   $\frac{1}{y^{3}}=1+C\left(1+t^{2}\right)\quad\Longrightarrow\quad y=\left[1+C\left(1+t^{2}\right)\right]^{-1/3}$

   Checking numerically with $C=0.7$ at $t=1.7$: the left side $3(1+t^2)y'=-1.60448567$ and the right side $2ty(y^3-1)=-1.60448567$. $\checkmark$ ($C=0$ recovers the equilibrium $y\equiv1$.)