Mister exam

# Differential equation dy/dx=(x+y+4)/(x-y-6)

Differential equation
with unknown function ()

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

from to

### The solution

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d           4 + x + y(x)
--(y(x)) = -------------
dx         -6 + x - y(x)
$$\frac{d}{d x} y{\left(x \right)} = \frac{x + y{\left(x \right)} + 4}{x - y{\left(x \right)} - 6}$$
y' = (x + y + 4)/(x - y - 6)
Detail solution
Given the equation:
$$\frac{d}{d x} y{\left(x \right)} - \frac{x + y{\left(x \right)} + 4}{x - y{\left(x \right)} - 6} = 0$$
Do replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)} + 5}{x - 1}$$
and because
$$y{\left(x \right)} = \left(x - 1\right) u{\left(x \right)} - 5$$
then
$$\frac{d}{d x} y{\left(x \right)} = \left(x - 1\right) \frac{d}{d x} u{\left(x \right)} + u{\left(x \right)}$$
substitute
$$- \frac{x}{x - \left(x - 1\right) u{\left(x \right)} - 1} + \left(x - 1\right) \frac{d}{d x} u{\left(x \right)} - \frac{\left(x - 1\right) u{\left(x \right)} - 5}{x - \left(x - 1\right) u{\left(x \right)} - 1} + u{\left(x \right)} - \frac{4}{x - \left(x - 1\right) u{\left(x \right)} - 1} = 0$$
or
$$\left(x - 1\right) \frac{d}{d x} u{\left(x \right)} + \frac{u^{2}{\left(x \right)} + 1}{u{\left(x \right)} - 1} = 0$$
This differential equation has the form:
f1(x)*g1(u)*u' = f2(x)*g2(u),

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

Divide both parts of the equation by f1(x)
$$x - 1$$
we get
$$\frac{d}{d x} u{\left(x \right)} = - \frac{u^{2}{\left(x \right)} + 1}{\left(x - 1\right) \left(u{\left(x \right)} - 1\right)}$$
Divide both parts of the equation by g2(u)
$$\frac{u^{2}{\left(x \right)} + 1}{u{\left(x \right)} - 1}$$
we get
$$\frac{\left(u{\left(x \right)} - 1\right) \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{1}{x - 1}$$
We separated the variables x and u.

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

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

do backward replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)} + 5}{x - 1}$$
$$\frac{\log{\left(1 + \frac{\left(y{\left(x \right)} + 5\right)^{2}}{\left(x - 1\right)^{2}} \right)}}{2} - \operatorname{atan}{\left(\frac{y{\left(x \right)} + 5}{x - 1} \right)} = Const - \log{\left(x - 1 \right)}$$
                      /      _________________\
|     /               2 |
|    /      (5 + y(x))  |       /5 + y(x)\
log(-1 + x) = C1 - log|   /   1 + ----------- | + atan|--------|
|  /                 2  |       \ -1 + x /
\\/          (-1 + x)   /                 
$$\log{\left(x - 1 \right)} = C_{1} - \log{\left(\sqrt{1 + \frac{\left(y{\left(x \right)} + 5\right)^{2}}{\left(x - 1\right)^{2}}} \right)} + \operatorname{atan}{\left(\frac{y{\left(x \right)} + 5}{x - 1} \right)}$$
Graph of the Cauchy problem
The classification
linear coefficients
1st power series
lie group
linear coefficients Integral
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 1.2833596014192565)
(-5.555555555555555, 1.4775028971649715)
(-3.333333333333333, 1.2880244904573492)
(-1.1111111111111107, 0.6039361991624052)
(1.1111111111111107, -0.9373729848374754)
(3.333333333333334, -3.6534115274805963)
(5.555555555555557, 6.234735095871012e-38)
(7.777777777777779, 8.388243567718494e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)
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