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3 years ago

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How many ordered pairs of positive integers satisfy xY=32?

1 answer:

IRINA_888 [86]3 years ago

4 0

Answer:

Step-by-step explanation:

xy = 32

x ; y is the divisor of 32

but : 32 = 2^5

6 ordered pairs of positive integers satisfy xY=32

(1;32) , (2;16) , (4; 8),(32; 1) , (16;2) , (8;4) .

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Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

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Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

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Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

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$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

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