AHL 5.14—Setting up a DE, solve by separating variables

$V$ is the volume of water at time $t$, and it is draining, meaning it decreases over time.

'Rate proportional to $V$' means the rate equals $k$ times $V$ for some constant $k>0$.

Since the tank is draining (decreasing), the rate of change must be negative.

Combining gives $\frac{dV}{dt}=-kV$.

$\frac{dV}{dt}=kV$ models growth, not draining — missing the negative sign.

$\frac{dV}{dt}=k(V-t)$ incorrectly mixes $t$ into the proportionality; the text only says proportional to $V$.

$\frac{dV}{dt}=-kt$ makes the rate proportional to time, not to the volume remaining — wrong quantity.

Only $\frac{dV}{dt}=-kV$ captures both proportionality to $V$ and the decreasing behaviour.

A DE is separable if the right-hand side can be written as $f(x)\cdot g(y)$, a product not a sum.

$x^2+y$ is a sum of a function of $x$ and a function of $y$, not a product.

There is no way to factor $x^2+y$ into $f(x)\cdot g(y)$ form.

Therefore $\frac{dy}{dx}=x^2+y$ is not separable; it needs another method such as an integrating factor.

$xy=f(x)\cdot g(y)$ with $f(x)=x$, $g(y)=y$, so it is separable.

$\frac{y}{x}=f(x)\cdot g(y)$ with $f(x)=\frac{1}{x}$, $g(y)=y$, so it is separable.

$3y=f(x)\cdot g(y)$ with $f(x)=3$, $g(y)=y$, so it is separable.

Only option B has $x$ and $y$ combined by addition, making it not separable.