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antoniya [11.8K]
3 years ago
11

I'm justing starting geometry honors, and I would like to know how to pass this class?! Is it easy or hard? HELP!!

Mathematics
2 answers:
rosijanka [135]3 years ago
8 0
Its different for different people but for me it was easy.
a_sh-v [17]3 years ago
5 0
I found geometry to be very easy it is a lot of fun brain teasers if u CAN do logic puzzles u will do well and enjoy it if you have a good teacher
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If the graph of y=f(x) passes through the point (0,1), and dy/dx=xsin(x^2)/y, then f(x)= ?
Zigmanuir [339]

The differential equation

d<em>y</em>/d<em>x</em> = <em>x</em> sin(<em>x</em> ²) / <em>y</em>

is separable as

<em>y</em> d<em>y</em> = <em>x</em> sin(<em>x</em> ²) d<em>x</em>

Integrate both sides:

∫ <em>y</em> d<em>y</em> = ∫ <em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ 2<em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ sin(<em>x</em> ²) d(<em>x</em> ²)

1/2 <em>y</em> ² = -1/2 cos(<em>x</em> ²) + <em>C</em>

Solve for <em>y</em> implicitly:

<em>y</em> ² = -cos(<em>x</em> ²) + <em>C</em>

Given that <em>y</em> = 1 when <em>x</em> = 0, we get

1² = -cos(0²) + <em>C</em>

1 = -cos(0) + <em>C</em>

1 = -1 + <em>C</em>

<em>C</em> = 2

Then the particular solution to the DE is

<em>y</em> ² = 2 - cos(<em>x</em> ²)

Solving explicitly for <em>y</em> would give two solutions,

<em>y</em> = ± √(2 - cos(<em>x</em> ²))

but only the one with the positive square root satisfies the initial condition:

√(2 - cos(0²)) = √1 = 1

-√(2 - cos(0²)) = -√1 = -1 ≠ 1

So the unique solution to this DE is

<em>y</em> = √(2 - cos(<em>x</em> ²))

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