# Differential equation (1+2y)xdx+(1+x^2)dy=0

y() =
y'() =
y''() =
y'''() =
y''''() =

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### The solution

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     2 d                     d
x + x *--(y(x)) + 2*x*y(x) + --(y(x)) = 0
dx                    dx          
$$x^{2} \frac{d}{d x} y{\left(x \right)} + 2 x y{\left(x \right)} + x + \frac{d}{d x} y{\left(x \right)} = 0$$
x^2*y' + 2*x*y + x + y' = 0
Detail solution
Given the equation:
$$x^{2} \frac{d}{d x} y{\left(x \right)} + 2 x y{\left(x \right)} + x + \frac{d}{d x} y{\left(x \right)} = 0$$
This differential equation has the form:
$$f_1(x)*g_1(y)*y' = f_2(x)*g_2(y)$$
where
$$f_{1}{\left(x \right)} = 1$$
$$g_{1}{\left(y \right)} = 1$$
$$f_{2}{\left(x \right)} = - \frac{x}{x^{2} + 1}$$
$$g_{2}{\left(y \right)} = 2 y{\left(x \right)} + 1$$
We give the equation to the form:
$$\frac{g_1(y)}{g_2(y)}*y'= \frac{f_2(x)}{f_1(x)}$$
Divide both parts of the equation by $g_{2}{\left(y{\left(x \right)} \right)}$
$$2 y{\left(x \right)} + 1$$
we get
$$\frac{\frac{d}{d x} y{\left(x \right)}}{2 y{\left(x \right)} + 1} = - \frac{x}{x^{2} + 1}$$
We separated the variables x and y.

Now, multiply the both equation sides by dx,
then the equation will be the
$$\frac{dx \frac{d}{d x} y{\left(x \right)}}{2 y{\left(x \right)} + 1} = - \frac{dx x}{x^{2} + 1}$$
or
$$\frac{dy}{2 y{\left(x \right)} + 1} = - \frac{dx x}{x^{2} + 1}$$

Take the integrals from the both equation sides:
- the integral of the left side by y,
- the integral of the right side by x.
$$\int \frac{1}{2 y + 1}\, dy = \int \left(- \frac{x}{x^{2} + 1}\right)\, dx$$
Detailed solution of the integral with y
Detailed solution of the integral with x
Take this integrals
$$\frac{\log{\left(2 y + 1 \right)}}{2} = Const - \frac{\log{\left(x^{2} + 1 \right)}}{2}$$
Detailed solution of the equation
We get the simple equation with the unknown variable y.
(Const - it is a constant)

Solution is:
$$y_{1} = y{\left(x \right)} = \frac{- x^{2} + C_{1}}{2 \left(x^{2} + 1\right)}$$
              2
C1 - x
y(x) = ----------
/     2\
2*\1 + x /
$$y{\left(x \right)} = \frac{- x^{2} + C_{1}}{2 \left(x^{2} + 1\right)}$$
$$y\left(x\right)={\it ilt}\left(-{{g_{19164}^3\,\left({{d^2}\over{d \,g_{19164}^2}}\,\mathcal{L}\left(y\left(x\right) , x , g_{19164} \right)\right)-y\left(0\right)\,g_{19164}^2+1}\over{g_{19164}^3}} , g_{19164} , x\right)$$
y = 'ilt(-(g19164^3*'diff('laplace(y,x,g19164),g19164,2)-y(0)*g19164^2+1)/g19164^3,g19164,x)
Graph of the Cauchy problem
The classification
1st exact
1st exact Integral
1st linear
1st linear Integral
1st power series
Bernoulli
Bernoulli Integral
almost linear
almost linear Integral
factorable
lie group
separable
separable Integral
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 1.55305170393819)
(-5.555555555555555, 3.462127623555846)
(-3.333333333333333, 9.924314850059593)
(-1.1111111111111107, 55.99863122678314)
(1.1111111111111107, 55.99863624207655)
(3.333333333333334, 9.924314673831544)
(5.555555555555557, 3.46212835124426)
(7.777777777777779, 1.553052294106616)
(10.0, 0.7500004440074922)
(10.0, 0.7500004440074922)
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