Mister Exam

Differential equation dy/dx=xy^(1/2)

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

from to

The solution

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d              ______
--(y(x)) = x*\/ y(x)
dx                   
$$\frac{d}{d x} y{\left(x \right)} = x \sqrt{y{\left(x \right)}}$$
y' = x*sqrt(y)
Detail solution
Given the equation:
$$\frac{d}{d x} y{\left(x \right)} = x \sqrt{y{\left(x \right)}}$$
This differential equation has the form:
f1(x)*g1(y)*y' = f2(x)*g2(y),

where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = x$$
$$\operatorname{g_{2}}{\left(y \right)} = \sqrt{y{\left(x \right)}}$$
We give the equation to the form:
g1(y)/g2(y)*y'= f2(x)/f1(x).

Divide both parts of the equation by g2(y)
$$\sqrt{y{\left(x \right)}}$$
we get
$$\frac{\frac{d}{d x} y{\left(x \right)}}{\sqrt{y{\left(x \right)}}} = x$$
We separated the variables x and y.

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

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

Solution is:
$$\operatorname{y_{1}} = y{\left(x \right)} = \frac{C_{1}^{2}}{4} + \frac{C_{1} x^{2}}{4} + \frac{x^{4}}{16}$$
         2    4       2
C1    x    C1*x
y(x) = --- + -- + -----
4    16     4  
$$y{\left(x \right)} = \frac{C_{1}^{2}}{4} + \frac{C_{1} x^{2}}{4} + \frac{x^{4}}{16}$$
The classification
separable
1st exact
Bernoulli
1st power series
lie group
separable Integral
1st exact Integral
Bernoulli Integral
The graph
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