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3 years ago

14

One part of 10 equal parts of a whole is called a

1 answer:

galina1969 [7]3 years ago

4 0

Answer:

tenth

Step-by-step explanation:

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(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to <em>x</em> :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to <em>y</em> :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

52 lbs is equal to how many kg

Butoxors [25]

1 lb = 0.45 kg
52 lb= 0.45 * 52
52 lb= 23.4 kg

4 0

3 years ago

Read 2 more answers

Help me for brainliest...

spin [16.1K]

Answer:

1cm = 320 km?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If RST = NPQ, then rt is congruent to<br> Α. NP<br> B. NQ<br> C. PQ<br> D. Qp

PIT_PIT [208]

NQ
This is because you have to see what R is congruent to (based off the order) and you see N. Then you do the same thing for T and you get Q.

4 0

3 years ago

Read 2 more answers

Find the values of x y and z

andrey2020 [161]

9514 1404 393

Answer:

x = 18√2
y = 18√3
z = 36

Step-by-step explanation:

These "special" right triangles have side length ratios that it is useful to remember.

<u>45°-45°-90° triangle</u>

Sides have the ratios 1 : 1 : √2. That is, x is √2 times as long as the side length shown as 18.

x = 18√2

<u>30°-60°-90° triangle</u>

Shortest to longest, sides have the ratios ...

1 : √3 : 2

That is, y is √3 times the length of the side marked 18, and z is 2 times the length of the side marked 18.

y = 18√3

z = 2·18 = 36

4 0

3 years ago