Verify that the given differential equation is not exact. (−xy sin(x) + 2y cos(x)) dx + 2x cos(x) dy = 0 if the given de is written in the form m(x, y) dx + n(x, y) dy = 0, one has my = nx = . since my and nx equal, the equation is not exact. multiply the given differential equation by the integrating factor μ(x, y) = xy and verify that the new equation is exact. if the new de is written in the form m(x, y) dx + n(x, y) dy = 0, one has my = nx = . since my and nx equal, the equation is exact. solve.

We are to verify and confirm if the given differential equations are exact or not. Then solve for the exact equation.

The first differential equation says:

Recall that:

A differential equation that takes the form will be exact if and only if:

From equation (1), we can represent M and N as follows:

Thus, taking the differential of M and N, we have:

From above, it is clear that:

∴

We can conclude that the equation is not exact.

Now, after multiplying the given differential equation in (1) by the integrating factor μ(x, y) = xy, we have:

Representing the equation into form M and N, then:

Taking the differential, we have:

Here;

Therefore, we can conclude that the second equation is exact.

Now, the solution of the second equation is as follows:

   ---- (3)

Taking integrations by parts:

∴

  ----- replace this equation into (3)

∴

Therefore, from the explanation, we've can conclude that the first equation is not exact and the second equation is exact.

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The ODE

is exact if

We have

so the ODE is indeed not exact.

Multiplying both sides of the ODE by gives

so that , and the modified ODE is exact.

We're looking for a solution of the form

so that by differentiation, we should have

Integrating both sides of the second equation with respect to gives

Differentiating both sides with respect to gives

for some constant .

So the general solution to this ODE is

or simply