Question

solve it by standard rules for finding I.F. (2xy4ey+2xy3+y)dx+(x2yey-x2y2-3x)​

Answer 1

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

Answer 2

Answer:

PLZ MARK AS BRAIN LISTS ANSWER

Step-by-step explanation:

Given an equation M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0, an integrating factor is a function μ(x,y)μ(x,y) which—when multiplied by the equation—makes it exact. That is, is satisfies

∂(μM)∂y=∂(μN)∂x∂(μM)∂y=∂(μN)∂x

There are several methods of finding integrating factors; one way is to make a “lucky” (or educated) guess. (See the end of this answer for a non-“lucky guess” approach.) With M=2xy4ey+2xy3+yM=2xy4ey+2xy3+y and N=x2y4ey−x2y2−3xN=x2y4ey−x2y2−3x, we can see that ∂M∂y∂M∂y would include a term involving y3eyy3ey, but no such term would appear in ∂N∂x∂N∂x. Recognizing this, it seems it would be helpful to separate y4y4and eyey into two separate factors, which we can do by dividing both M and N by y4y4—that is, we might guess that y−4y−4 is (at least a factor of) μμ.