-2.2.12 IVP y'=2x/(y+x^2y), y(0)=-2

[karush]

693
2.2.13 (a) find initial value (b)plot and (c) interval

(a) find initial value (b)plot and (c) interval
$$\displaystyle
y^{\prime}=2x/(y+x^2y), \quad y(0)=-2$$
separate the variables
$$\frac{dy}{dx}=\frac{2x}{y+x^2y}=\frac{2x}{y(1-x^2)}$$
$$y\, dy =\frac{2x}{(1-x^2)}dx$$
integrate
$$\int y\, dy = -\sqrt{4}\int \frac{x}{(1-x^2)} \, dx$$
$$\frac{y^2}{2}= -2\frac{\ln(1 - x^2)}{2}$$
$$y=-\sqrt{4}\sqrt{\ln(1-x^2)}$$ok I am stuck again.,...book answer
$$(a)\, y = −[2 ln(1 + x^2) + 4]^{-1/2} \\ (c) −\infty < x <\infty$$

 

[skeeter]

$\dfrac{dy}{dx} = \dfrac{2x}{y(1+x^2)}$

$ y \, dy = \dfrac{2x}{1+x^2} \, dx$

$\dfrac{y^2}{2} = \ln(1+x^2) + C$

$y(0) = -2 \implies C = 2$ ...

$\dfrac{y^2}{2} = \ln(1+x^2) + 2$

$y = -\sqrt{2\ln(1+x^2) + 4} = -[2\ln(1+x^2)+4]^{1/2}$

... check the sign of the exponent on the book "answer"

 

[karush]

$\dfrac{dy}{dx} = \dfrac{2x}{y(1+x^2)}$

$ y \, dy = \dfrac{2x}{1+x^2} \, dx$

$\dfrac{y^2}{2} = \ln(1+x^2) + C$

$y(0) = -2 \implies C = 2$ ...

$\dfrac{y^2}{2} = \ln(1+x^2) + 2$

$y = -\sqrt{2\ln(1+x^2) + 4} = -[2\ln(1+x^2)+4]^{1/2}$

... check the sign of the exponent on the book "answer"

you are correct no negative exponent

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