Two integrating factors giving two different equations

Question

In the above question, we get two integrating factors 1/y² and t. These give rise to two different equations. Please explain the underlying phenomenon of why this happens.

Answer 1

If we use $y^{-2}$ as the integrating factor, we have, \begin{align*} yt \ \mathrm{d}y + \frac{1}{2} y^2 \ \mathrm{d}t = 0 \implies \frac{y^2 t}{2} = C. \end{align*} Similarly, if we use $t$ as the integrating factor, we have,\begin{align*} \frac{1}{4} \cdot \left(4y^3t^2 \ \mathrm{d}y + 2 y^4 t \ \mathrm{d}t \right) = 0 \implies \frac{y^4 t^2}{4} = C. \end{align*} So as long as $t > 0$, both equations yield exactly the same relation. Moreover, from the final result, we can conclude that $y^{-2}$ is the same as $t$ upto a constant factor which is why both of them are integrating factors.

Answer 2

Multiply the equation by $1/y^2$ and you obtain $$ ytdy +\frac{1}{2}y^2dt = \frac{1}{2}d(ty^2) = 0. $$ If you multiply it by $t$, you obtain $$ y^3t^2dy +\frac{1}{2}ty^4dt = \frac{1}{4}d(t^2y^4) = \frac{1}{2}ty^2d(ty^2) = 0. $$ More generally, if you multiply the equation by $\phi(y,t)/y^2$, where $\phi(y,t)$ is an arbitrary function of $y$ and $t$, you obtain $$ \frac{1}{2}\phi(y,t)d(ty^2)=0. $$ Although the equations are different, they have the same solution, namely $ty^2=C$.