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dy/dx=ycosx/1+2y^2

Differential equation dy/dx=ycosx/1+2y^2

For Cauchy problem:

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

The graph:

from to

The solution

You have entered [src]
d             2                 
--(y(x)) = 2*y (x) + cos(x)*y(x)
dx                              
$$\frac{d}{d x} y{\left(x \right)} = 2 y^{2}{\left(x \right)} + y{\left(x \right)} \cos{\left(x \right)}$$
y' = 2*y^2 + y*cos(x)
The answer [src]
              sin(x)       
             e             
y(x) = --------------------
                /          
               |           
               |  sin(x)   
       C1 - 2* | e       dx
               |           
              /            
$$y{\left(x \right)} = \frac{e^{\sin{\left(x \right)}}}{C_{1} - 2 \int e^{\sin{\left(x \right)}}\, dx}$$
Graph of the Cauchy problem
The classification
Bernoulli
1st power series
lie group
Bernoulli Integral
Numerical answer [src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 2546789.040729482)
(-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, 2.125757255287192e+160)
(7.777777777777779, 8.388243571812246e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)
The graph
Differential equation dy/dx=ycosx/1+2y^2
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