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

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• karush
In summary, we are given a differential equation $\displaystyle y^{\prime}=2x/(y+x^2y)$ with initial value $y(0)=-2$. We separate the variables and integrate to get the solution $y=-[2\ln(1+x^2)+4]^{1/2}$, with an interval of $-\infty<x<\infty$. However, we must also check the sign of the exponent in the book's solution, as it may be incorrect.
karush
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MHB
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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
$$(a)\, y = −[2 ln(1 + x^2) + 4]^{-1/2} \\ (c) −\infty < x <\infty$$

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

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

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

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

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

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

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

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

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

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

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

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

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

... check the sign of the exponent on the book "answer"
you are correct no negative exponent

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Last edited:

## 1. What is an IVP?

An IVP, or initial value problem, is a type of differential equation that involves finding a solution for a function with a given set of initial conditions. In this case, the function is y(x) and the initial condition is y(0)=-2.

## 2. What does y' mean in the given equation?

The notation y' represents the derivative of the function y with respect to x. In other words, it represents the rate of change of y with respect to x.

## 3. What is the role of x and y in the equation?

In this equation, x and y are both independent variables. The function y(x) represents the dependent variable, and the equation shows how y changes as x changes.

## 4. How do I solve this particular IVP?

To solve this IVP, you can use techniques such as separation of variables or substitution. By solving for y(x), you can find the particular solution that satisfies the given initial condition y(0)=-2.

## 5. What are the real-life applications of this type of IVP?

This type of IVP can be used to model various physical systems, such as population growth, chemical reactions, and radioactive decay. It can also be applied in economics and engineering to study changes in quantities over time.

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