Show whether the differential equation is exact or not. `sin(t)dy/dt + t^2 e^y dy/dt - dy/dt + ycos(t) + 2te^y − 3t^2 = 0`

Show whether the differential equation is exact or not.

`sin(t)dy/dt + t^2 e^y dy/dt - dy/dt + ycos(t) + 2te^y − 3t^2 = 0`

Solution:

To determine whether the given differential equation is exact or not, we need to check if it satisfies the condition of exactness.

A differential equation of the form `M(x, y) dx + N(x, y) dy = 0` is said to be exact if and only if the partial derivatives of M with respect to y and N with respect to x are equal, i.e., `{∂M}/{∂y} = {∂N}/{∂x}`.

Let's analyze the given differential equation: `sin(t)dy/dt + t^2 e^y dy/dt - dy/dt + ycos(t) + 2te^y − 3t^2 = 0`

Rearranging the equation, we have:

`(sin(t) + t^2 e^y - 1) dy/dt + (ycos(t) + 2te^y − 3t^2) = 0`

Comparing this with M(x, y) dx + N(x, y) dy = 0, we have:

`M(x, y) = sin(t) + t^2 e^y - 1`

`N(x, y) = ycos(t) + 2te^y − 3t^2`

Now, we need to calculate the partial derivatives:

`{∂M}/{∂y} = t^2 e^y`

`{∂N}/{∂x} = -sin(t) + 2te^y`

Since `{∂M}/{∂y}` is not equal to `{∂N}/{∂x}`, the given differential equation is not exact.

Therefore, the given differential equation is not exact.