Mister Exam

Differential equation xydx-(x+2y)^2dy=0

For Cauchy problem:

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

The graph:

from to

The solution

You have entered [src]
          2 d             2    d              d                
x*y(x) - x *--(y(x)) - 4*y (x)*--(y(x)) - 4*x*--(y(x))*y(x) = 0
            dx                 dx             dx               
$$- x^{2} \frac{d}{d x} y{\left(x \right)} - 4 x y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + x y{\left(x \right)} - 4 y^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
-x^2*y' - 4*x*y*y' + x*y - 4*y^2*y' = 0
Detail solution
Given the equation:
$$- x^{2} \frac{d}{d x} y{\left(x \right)} - 4 x y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + x y{\left(x \right)} - 4 y^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
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
$$- 4 x^{2} u^{2}{\left(x \right)} \frac{d}{d x} x u{\left(x \right)} - 4 x^{2} u{\left(x \right)} \frac{d}{d x} x u{\left(x \right)} + x^{2} u{\left(x \right)} - x^{2} \frac{d}{d x} x u{\left(x \right)} = 0$$
or
$$- 4 x^{3} u^{2}{\left(x \right)} \frac{d}{d x} u{\left(x \right)} - 4 x^{3} u{\left(x \right)} \frac{d}{d x} u{\left(x \right)} - x^{3} \frac{d}{d x} u{\left(x \right)} - 4 x^{2} u^{3}{\left(x \right)} - 4 x^{2} u^{2}{\left(x \right)} = 0$$
This differential equation has the form:
f1(x)*g1(u)*u' = f2(x)*g2(u),

where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(u \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x}$$
$$\operatorname{g_{2}}{\left(u \right)} = \frac{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}}{4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1}$$
We give the equation to the form:
g1(u)/g2(u)*u'= f2(x)/f1(x).

Divide both parts of the equation by g2(u)
$$\frac{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}}{4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1}$$
we get
$$\frac{\left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right) \frac{d}{d x} u{\left(x \right)}}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{1}{x}$$
We separated the variables x and u.

Now, multiply the both equation sides by dx,
then the equation will be the
$$\frac{dx \left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right) \frac{d}{d x} u{\left(x \right)}}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{dx}{x}$$
or
$$\frac{du \left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right)}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{dx}{x}$$

Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \frac{4 u^{2} + 4 u + 1}{4 u^{2} \left(u + 1\right)}\, du = \int \left(- \frac{1}{x}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$\frac{3 \log{\left(u \right)}}{4} + \frac{\log{\left(u + 1 \right)}}{4} - \frac{1}{4 u} = Const - \log{\left(x \right)}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)

Solution is:
$$\operatorname{u_{1}} = \log{\left(x \right)} + \frac{\log{\left(u{\left(x \right)} + 1 \right)}}{4} + \frac{3 \log{\left(u{\left(x \right)} \right)}}{4} - \frac{1}{4 u{\left(x \right)}} = C_{1}$$
do backward replacement
$$y{\left(x \right)} = x u{\left(x \right)}$$
$$y1 = y(x) = C_{1} x$$
The answer [src]
                    /    __________\         
                    |   /      x   |     x   
log(y(x)) = C1 - log|4 /  1 + ---- | + ------
                    \\/       y(x) /   4*y(x)
$$\log{\left(y{\left(x \right)} \right)} = C_{1} + \frac{x}{4 y{\left(x \right)}} - \log{\left(\sqrt[4]{\frac{x}{y{\left(x \right)}} + 1} \right)}$$
Graph of the Cauchy problem
The classification
factorable
1st homogeneous coeff best
1st homogeneous coeff subs indep div dep
1st homogeneous coeff subs dep div indep
1st power series
lie group
1st homogeneous coeff subs indep div dep Integral
1st homogeneous coeff subs dep div indep Integral
Numerical answer [src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 0.5325491523151631)
(-5.555555555555555, 0.3415339359004146)
(-3.333333333333333, 0.1780323336116803)
(-1.1111111111111107, 0.04662848651693114)
(1.1111111111111107, -0.026636560043214227)
(3.333333333333334, -0.09018889769322533)
(5.555555555555557, -0.15995764155772052)
(7.777777777777779, -0.233871162973425)
(10.0, -0.3110317790429888)
(10.0, -0.3110317790429888)
The graph
Differential equation xydx-(x+2y)^2dy=0
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