Divisibility for 356

Anarel [89]

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—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

7 0

3 years ago

A bathtub is draining at a constant rate. After 2 minutes, it holds 30 gallons of water. Three minutes later, it holds 12 gallon

storchak [24]

The answer is boxed on the side

5 0

3 years ago

Read this E[2X^2 â€" Y].

djyliett [7]

Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2X ² - Y.

If you don't know how X or Y are distributed, but you know E[X ²] and E[Y], then it's as simple as distributing the expectation over the sum:

E[2X ² - Y] = 2 E[X ²] - E[Y]

Or, if you're given the expectation and variance of X, you have

Var[X] = E[X ²] - E[X]²

→ E[2X ² - Y] = 2 (Var[X] + E[X]²) - E[Y]

Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.

6 0

3 years ago

If two segments from the same exterior point are tangent to the circle then the two segment are congruent

Dimas [21]

This seems like a true or false question this question is true because they haven't changed in length or size at all

3 0

3 years ago

Can somebody help, will give brainliest

grigory [225]

9514 1404 393

Answer:

c = 2
segments 10, 8, 6 (top down)

Step-by-step explanation:

The sum of top and bottom bases is twice the midsegment.

(c² +6) +(c² +2) = 2(4c)

2c² -8c +8 = 0 . . . . . . . . . subtract 8c

2(c -2)² = 0 . . . . . . . . factor

c = 2 . . . . . . the value that makes the binomial be zero

The value of c is 2; the segment lengths (top-down) are 10, 8, 6.

5 0

3 years ago