Question #4a5e2

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Question #4a5e2

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Answer 1 · 2017-04-18T17:34:26

$y = - \frac{2}{x^{2} - 2}$ $y = -2/(x^2- 2)$

Explanation:

Given: $\frac{d y}{d x} = x y^{2}; y (0) = 1$ $dy/dx = xy^2; y(0)=1$

Use the separation of variable method:

$\frac{d y}{y^{2}} = x d x$ $dy/y^2= xdx$

Integrate:

$\int \frac{d y}{y^{2}} = \int x d x$ $intdy/y^2= intxdx$

$- \frac{1}{y} = \frac{x^{2}}{2} + C$ $-1/y = x^2/2+C$

$- \frac{2}{y} = x^{2} + C$ $-2/y = x^2+C$

$y = - \frac{2}{x^{2} + C}$ $y = -2/(x^2+ C)$

Evaluate at the boundary condition, $y (0) = 1$ $y(0) = 1$ :

$1 = - \frac{2}{0^{2} + C}$ $1 = -2/(0^2+C)$

$C = - 2$ $C =-2$

The equation is:

$y = - \frac{2}{x^{2} - 2}$ $y = -2/(x^2- 2)$

Answer 2 · 2017-04-18T17:41:01

$y = \frac{2}{2 - x^{2}}$ $y = 2/(2 -x^2)$

Explanation:

$\frac{d y}{d x} = x y^{2}$ $(dy)/(dx) = xy^2$

$\frac{1}{y^{2}} d y = x d x$ $1/y^2 dy= x dx$

$\int \frac{1}{y^{2}} d y = \int x d x$ $int 1/y^2 dy= int x dx$

$- \frac{1}{y} = \frac{x^{2}}{2} + c$ $-1/y = x^2/2 +c$

plug in $y = 1 and x = 0$ $y = 1 and x =0$ in the above equation

$- \frac{1}{1} = \frac{0^{2}}{2} + c$ $-1/1 = 0^2/2 +c$ $\to c = - 1$ $-> c =-1$

therefore,
$- \frac{1}{y} = \frac{x^{2}}{2} - 1 = \frac{x^{2} - 2}{2}$ $-1/y = x^2/2 -1 = (x^2 - 2)/2$

$- \frac{2}{x^{2} - 2} = y$ $-2/(x^2 - 2) = y$

$\frac{2}{2 - x^{2}} = y$ $2/(2 -x^2) = y$

Answer 3 · 2017-04-18T17:41:50

$y = \frac{2}{2 - x^{2}}$ $y=2/(2-x^2)$

Explanation:

$\frac{d y}{d x} = x y^{2}$ $dy/dx=xy^2$

$\int y^{- 2}$ $inty^-2$ $d y = \int x$ $dy=intx$ $d x$ $dx$

$- \frac{1}{y} = \frac{1}{2} x^{2} + c$ $-1/y=1/2x^2+"c"$

Given that $y = 1, x = 0$ $y=1, x=0$ ,

$- \frac{1}{1} = \frac{1}{2} {(0)}^{2} + c \Rightarrow c = - 1$ $-1/1=1/2(0)^2+"c" rArr"c"=-1$

Since they haven't asked to give it in the form $y = f (x)$ $y=f(x)$ , you can leave it like this. But I will solve it for $y$ $y$ just in case you would like to see how to do that.

$- \frac{1}{y} = \frac{1}{2} x^{2} - 1$ $-1/y=1/2x^2-1$

$\frac{1}{y} = 1 - \frac{1}{2} x^{2} = \frac{1}{2} (2 - x^{2})$ $1/y=1-1/2x^2=1/2(2-x^2)$

$y = {(\frac{1}{2} (2 - x^{2}))}^{- 1}$ $y=(1/2(2-x^2))^-1$

$y = \frac{2}{2 - x^{2}}$ $y=2/(2-x^2)$

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