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Likurg_2 [28]
3 years ago
15

We are given a differential equation and will rewrite it in the form M(x, y) dx + N(x, y) dy = 0. xy2 dy dx = y3 − x3 xy2 dy = (

y3 − x3) dx xy2 dy − (y3 − x3) dx = 0 xy2 dy + (−y3 + x3) dx = 0
Mathematics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

Check attachment for orderliness of question

Step-by-step explanation:

We want to the equation in an exact form

M(x, y) dx + N(x, y) dy = 0.

xy² dy/dx = y³− x³

Cross multiply

Then,

xy²dy=(y³-x³)dx

xy²dy - (y³-x³)dx = 0

xy²dy + (x³-y³)dx=0

Therefore,

(x³-y³)dx + xy²dy=0

Can only be an exact equation if and only if

dM/dy=dN/dx

From the equation

M=(x³-y³)

Then, dM/dy=-3y²

Also, N=xy²

dN/dx=y²

Since dM/dy ≠ dN/dx

Then it is not an examct equation and we can say

M≠(x³-y³)

And

N≠(xy²)

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Answer:

Type 1 error:

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Type 2 error:

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When something that is true, is been rejected, then it's reffered to as Type I, on the other hand when

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Type II ;

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