Given the function $f(x,y) = ax^2y + by + ct^2 $ with $x(t)=2t$ and $y(t)=3t^2$ I shall determine the coefficients a,b,c so that $f(x,y)$ is a conserved quantity. My approach is $$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} = 0$$ In the end, after substituting $x$ and $y$, I get: $$24at^3 + 6tb + ct = 0$$ How do I continue by now? $a=b=c=0$ probably isn't the intented solution. Thanks


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as you say you need that $$ 24at^3 + 3tb + ct = 0 $$ (with a small correction to your calculation)

Which for $t \neq 0 $ is the same that $$ 24at^2 + 3b + c = 0 $$ for all no zero $t$. But if $a \neq 0$ you have a polynomial in $t$, which have almost two root for given coeficients, so the equality can't hold for all t, which force us to take $a = 0$. Now we have: $$ 3b +c = 0 \implies c=-3b $$ So $f$ is a conserved quantity if is of the form $$ f(x,y) = by - 3 b t^2 $$


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