Mister Exam

Differential equation sec^2x*tgy*dy+sec^2ytgxdx=0

For Cauchy problem:

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

The graph:

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The solution

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   2                   2    d                     
sec (y(x))*tan(x) + sec (x)*--(y(x))*tan(y(x)) = 0
$$\tan{\left(x \right)} \sec^{2}{\left(y{\left(x \right)} \right)} + \tan{\left(y{\left(x \right)} \right)} \sec^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
tan(x)*sec(y)^2 + tan(y)*sec(x)^2*y' = 0
Detail solution
Given the equation:
$$\tan{\left(x \right)} \sec^{2}{\left(y{\left(x \right)} \right)} + \tan{\left(y{\left(x \right)} \right)} \sec^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
This differential equation has the form:
f1(x)*g1(y)*y' = f2(x)*g2(y),

$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \sin{\left(2 x \right)}$$
$$\operatorname{g_{2}}{\left(y \right)} = \frac{1}{\sin{\left(2 y{\left(x \right)} \right)}}$$
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)
$$\frac{1}{\sin{\left(2 y{\left(x \right)} \right)}}$$
we get
$$\sin{\left(2 y{\left(x \right)} \right)} \frac{d}{d x} y{\left(x \right)} = - \sin{\left(2 x \right)}$$
We separated the variables x and y.

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

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

Solution is:
$$\operatorname{y_{1}} = y{\left(x \right)} = \pi - \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$
$$\operatorname{y_{2}} = y{\left(x \right)} = \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$
The answer [src]
            acos(C1 - cos(2*x))
y(x) = pi - -------------------
$$y{\left(x \right)} = \pi - \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$
       acos(C1 - cos(2*x))
y(x) = -------------------
$$y{\left(x \right)} = \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$
Graph of the Cauchy problem
The classification
1st power series
lie group
separable Integral
Numerical answer [src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 5.368083552329663e-09)
(-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, 7.321204927535585e+169)
(7.777777777777779, 8.38824356695596e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)
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