Question

Solve The Non Exact differntial Equation dx+(x/y-sin(y))dy=0.​

Answer 1

Answer:

(3x2y3 − 5x4) dx + (y + 3x3y2) dy = 0

In this case we have:

M(x, y) = 3x2y3 − 5x4

N(x, y) = y + 3x3y2

We evaluate the partial derivatives to check for exactness.

∂M∂y = 9x2y2

∂N∂x = 9x2y2

They are the same! So our equation is exact.

We can proceed.

Now we want to discover I(x, y)

Let's do the integration with x as an independent variable:

I(x, y) = ∫M(x, y) dx

= ∫(3x2y3 − 5x4) dx

= x3y3 − x5 + f(y)

Note: f(y) is our version of the constant of integration "C" because (due to the partial derivative) we had y as a fixed parameter that we know is really a variable.

So now we need to discover f(y)

At the very start of this page we said that N(x, y) can be replaced by ∂I∂y , so:

∂I∂y = N(x, y)

Which gets us:

3x3y2 + dfdy = y + 3x3y2

Cancelling terms:

dfdy = y

Integrating both sides:

f(y) = y22 + C

We have f(y). Now just put it in place:

I(x, y) = x3y3 − x5 + y22 + C

and the general solution (as mentioned before this example) is:

I(x, y) = C

Ooops! That "C" can be a different value to the "C" just before. But they both mean "any constant", so let's call them C1 and C2 and then roll them into the C below by saying C=C1+C2

So we get:

x3y3 − x5 + y22 = C

And that's how this method works!

Since that was our first example, let's go further and make sure our solution is correct.

Let's derive I(x, y) with respect to x, that is:

Evaluate ∂I∂x

Start with:

I(x, y) = x3y3 − x5 + y22

Using implicit differentiation we get

∂I∂x = x33y2y' + 3x2y3 − 5x4 + yy'

Simplify

∂I∂x = 3x2y3 − 5x4 + y'(y + 3x3y2)

We use the facts that y' = dydx and ∂I∂x = 0, then multiply everything by dx to finally get:

(y + 3x3y2)dy + (3x2y3 − 5x4)dx = 0

which is our original differential equation.

And so we know our solution is correct.

Step-by-step explanation:

Hope this will help you