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karush

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(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

$$(a)\, y = −[2 ln(1 + x^2) + 4]^{-1/2} \\ (c) −\infty < x <\infty$$

**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

$$(a)\, y = −[2 ln(1 + x^2) + 4]^{-1/2} \\ (c) −\infty < x <\infty$$

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