Show that the differential equation `2ty dy/dt + 2t + y^2 = 0`,
is the total derivative of the potential function `φ(t, y(t)) = t^2 + ty^2`

Solution:

To show that the given differential equation `2ty dy/dt + 2t + y^2 = 0` is the total derivative of the potential function `φ(t, y(t)) = t^2 + ty^2`, we need to verify if the partial derivatives of φ with respect to t and y satisfy the given equation.

Let's calculate the partial derivatives of φ(t, y) with respect to t and y:

`{∂φ}/{∂t} = 2t + y^2`

`{∂φ}/{∂y} = 2ty`

Now, taking the total derivative of φ(t, y) with respect to t:

`{dφ}/{dt} = {∂φ}/{∂t} dt + {∂φ}/{∂y} {dy}/{dt}`

Substituting the partial derivatives we calculated earlier:

`{dφ}/{dt} = (2t + y^2) dt + (2ty) {dy}/{dt}`

Comparing this with the given differential equation `2ty {dy}/{dt} + 2t + y^2 = 0`, we can see that they match:

`{dφ}/{dt} = 2ty {dy}/{dt} + 2t + y^2`

Hence, we have shown that the given differential equation is the total derivative of the potential function `φ(t, y(t)) = t^2 + ty^2`.