Non linear 2nd order ode not able to solve

[varen90]

u'u''-k1u=-a*cos(hy)-b
where,u'=du/dy;
and and a,b,k1 are constants
conditions
u(-H)=0;
u'(0)=0;
where 2H is height of the channel where the liquid is flowing
please help any suggestions are welcome
i couldn't find the analytical soln
numerical soln also am havin a dead end so please

 

[MathematicalPhysicist]

u'u''-k1u=-a*cos(hy)-b

(u')^2 u'' - k1 uu' =(-a cos(hy) -b)u'

1/3((u')^3)' - k1/2 ((u)^2)' = (-a cos(hy) -b) u'

Integrate both sides wrt y, where the RHS is:
$$\int_{-H}^{y} (-a \cdot cos(hy) -b) u' = (-a \cdot cos(hy) -b) u(y) - \int (ah \cdot sin(hy) u(y)$$

It doesn't look like there's an analytic solution.

You should try solve it by using numerical methods for solving integral equations.

I can't help more than this, sorry.

 

[varen90]

thx for the input really appreciated