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Charra [1.4K]
3 years ago
6

How do you simplify?

Mathematics
1 answer:
cupoosta [38]3 years ago
4 0
5. -40
6. -1/64 or in decimal form (<span>−0.015625</span>)
8. -25
9. 4/5
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perimeter of larger hexagon=2.5 perimeter of smaller hexagon

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What is the value of x?
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Dy/dx = 2xy^2 and y(-1) = 2 find y(2)
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2887301

—————

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}\\\\&#10;\mathsf{y^{-2}\,dy=2x\,dx}


Integrate both sides:

\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\&#10;\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\&#10;\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\&#10;\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}}\\\\\\&#10;\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\&#10;\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\&#10;\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\&#10;\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}}}\\\\\\&#10;\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\&#10;\mathsf{y=-\,\dfrac{2}{2x^2-3}}


So  y(– 2)  is

\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

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