# Question 4a5e2

Apr 18, 2017

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

#### Explanation:

Given: dy/dx = xy^2; y(0)=1#

Use the separation of variable method:

$\frac{\mathrm{dy}}{y} ^ 2 = x \mathrm{dx}$

Integrate:

$\int \frac{\mathrm{dy}}{y} ^ 2 = \int x \mathrm{dx}$

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

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

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

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

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

$C = - 2$

The equation is:

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

Apr 18, 2017

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x {y}^{2}$

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

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

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

plug in $y = 1 \mathmr{and} x = 0$ in the above equation

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

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

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

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

Apr 18, 2017

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x {y}^{2}$

$\int {y}^{-} 2$ $\mathrm{dy} = \int x$ $\mathrm{dx}$

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

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

$- \frac{1}{1} = \frac{1}{2} {\left(0\right)}^{2} + \text{c" rArr"c} = - 1$

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

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

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

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

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