# Differential equation dy/dx=(x^2+3y^2)/(2xy)

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

from to

### The solution

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            2      2
d          x  + 3*y (x)
--(y(x)) = ------------
dx           2*x*y(x)  
$$\frac{d}{d x} y{\left(x \right)} = \frac{x^{2} + 3 y^{2}{\left(x \right)}}{2 x y{\left(x \right)}}$$
y' = (x^2 + 3*y^2)/(2*x*y)
Detail solution
Given the equation:
$$\frac{d}{d x} y{\left(x \right)} - \frac{x^{2} + 3 y^{2}{\left(x \right)}}{2 x y{\left(x \right)}} = 0$$
Do replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)}}{x}$$
and because
$$y{\left(x \right)} = x u{\left(x \right)}$$
then
$$\frac{d}{d x} y{\left(x \right)} = x \frac{d}{d x} u{\left(x \right)} + u{\left(x \right)}$$
substitute
$$- \frac{3 u{\left(x \right)}}{2} + \frac{d}{d x} x u{\left(x \right)} - \frac{1}{2 u{\left(x \right)}} = 0$$
or
$$x \frac{d}{d x} u{\left(x \right)} - \frac{u{\left(x \right)}}{2} - \frac{1}{2 u{\left(x \right)}} = 0$$
This differential equation has the form:
$$f_1(x)*g_1(u)*u' = f_2(x)*g_2(u)$$
where
$$f_{1}{\left(x \right)} = 1$$
$$g_{1}{\left(u \right)} = 1$$
$$f_{2}{\left(x \right)} = - \frac{1}{x}$$
$$g_{2}{\left(u \right)} = - \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
We give the equation to the form:
$$\frac{g_1(u)}{g_2(u)}*u'= \frac{f_2(x)}{f_1(x)}$$
Divide both parts of the equation by g_2(u)
$$- \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
we get
$$- \frac{2 u{\left(x \right)} \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{1}{x}$$
We separated the variables x and u.

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

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

Solution is:
$$u_{1} = u{\left(x \right)} = - \sqrt{C_{1} x - 1}$$
$$u_{2} = u{\left(x \right)} = \sqrt{C_{1} x - 1}$$
do backward replacement
$$y{\left(x \right)} = x u{\left(x \right)}$$
$$y_{1} = y{\left(x \right)} = - x \sqrt{C_{1} x - 1}$$
$$y_{2} = y{\left(x \right)} = x \sqrt{C_{1} x - 1}$$
            ___________
y(x) = -x*\/ -1 + C1*x 
$$y{\left(x \right)} = - x \sqrt{C_{1} x - 1}$$
           ___________
y(x) = x*\/ -1 + C1*x 
$$y{\left(x \right)} = x \sqrt{C_{1} x - 1}$$
$$y\left(x\right)={\it ilt}\left({{\mathcal{L}\left({{3\,y\left(x \right)}\over{2\,x}}+{{x}\over{2\,y\left(x\right)}} , x , g_{19164} \right)+y\left(0\right)}\over{g_{19164}}} , g_{19164} , x\right)$$
y = 'ilt(('laplace((3*y)/(2*x)+x/(2*y),x,g19164)+y(0))/g19164,g19164,x)
Graph of the Cauchy problem
The classification
1st homogeneous coeff best
1st homogeneous coeff subs dep div indep
1st homogeneous coeff subs dep div indep Integral
1st homogeneous coeff subs indep div dep
1st homogeneous coeff subs indep div dep Integral
Bernoulli
Bernoulli Integral
factorable
lie group
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 4.025199483059684e-10)
(-5.555555555555555, 2.17e-322)
(-3.333333333333333, nan)
(-1.1111111111111107, 2.78363573e-315)
(1.1111111111111107, 8.427456047434801e+197)
(3.333333333333334, 3.1933833808213433e-248)
(5.555555555555557, 1.0400058166471786e-42)
(7.777777777777779, 8.388243567338858e+296)
(10.0, 9.527767843605595e-305)
(10.0, 9.527767843605595e-305)
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share to all your student friends: