Answer:
D is the answer
Step-by-step explanation:
<h3>
Answer: x = 7 and y = 3</h3>
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Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
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Let's consider case A
We have this system of equations

Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
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Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
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Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.
First, you must know these formula d(e^f(x) = f'(x)e^x dx, e^a+b=e^a.e^b, and d(sinx) = cosxdx, secx = 1/ cosx
(secx)dy/dx=e^(y+sinx), implies <span>dy/dx=cosx .e^(y+sinx), and then
</span>dy=cosx .e^(y+sinx).dx, integdy=integ(cosx .e^(y+sinx).dx, equivalent of
integdy=integ(cosx .e^y.e^sinx)dx, integdy=e^y.integ.(cosx e^sinx)dx, but we know that d(e^sinx) =cosx e^sinx dx,
so integ.d(e^sinx) =integ.cosx e^sinx dx,
and e^sinx + C=integ.cosx e^sinxdx
finally, integdy=e^y.integ.(cosx e^sinx)dx=e^2. (e^sinx) +C
the answer is
y = e^2. (e^sinx) +C, you can check this answer to calculate dy/dx