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Ecuación diferencial yy'+y^2=cosx

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

Ha introducido [src]
 2      d                     
y (x) + --(y(x))*y(x) = cos(x)
        dx                    
$$y^{2}{\left(x \right)} + y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = \cos{\left(x \right)}$$
y^2 + y*y' = cos(x)
Respuesta [src]
           __________________________________ 
          /                             -2*x  
       -\/  10*sin(x) + 20*cos(x) + C1*e      
y(x) = ---------------------------------------
                          5                   
$$y{\left(x \right)} = - \frac{\sqrt{C_{1} e^{- 2 x} + 10 \sin{\left(x \right)} + 20 \cos{\left(x \right)}}}{5}$$
          __________________________________
         /                             -2*x 
       \/  10*sin(x) + 20*cos(x) + C1*e     
y(x) = -------------------------------------
                         5                  
$$y{\left(x \right)} = \frac{\sqrt{C_{1} e^{- 2 x} + 10 \sin{\left(x \right)} + 20 \cos{\left(x \right)}}}{5}$$
Clasificación
Bernoulli
almost linear
1st power series
lie group
Bernoulli Integral
almost linear Integral