Differential equation y''-a(y'^2)-b(siny-a*cosy)=0

[mst_ab]

solve this differential equation:

y''-a(y'^2)-b(siny-a*cosy)=0

(a&b are cte.)

 

[Mark44]

cte. - ??

 

[gabbagabbahey]

I'm assuming that "cte." means constants?

The first step is to classify the differential equation: Is it ordinary? Is it linear? Is it homogeneous?

Once you classify it, think of which methods you know that can be used to solve this type of DE. Which methods come to mind in this case?

 

[mst_ab]

a&b are constant,
i found the way of solving this !:

v=y'
=> y''=v(dv/dy)

after solving, i get this equation:

y'^2=(-0.82b*cosy-0.9b*siny)/1.36

(a=0.3)

I'm not noob! I'm professor of university!

please help me to solve this!:
y'^2=(-0.82b*cosy-0.9b*siny)/1.36

 

[gabbagabbahey]

Umm...if y is function of x, then y''=v(dv/dx) not vdv/dy...What variable(s) is y actually a function of here?

 

[HallsofIvy]

Umm...if y is function of x, then y''=v(dv/dx) not vdv/dy...What variable(s) is y actually a function of here?

No, gabbagabbahey, that is a perfectly valid method. If v= dy/dx, then
$$\frac{d^2y}{dx^2}= \frac{dv}{dx}$$
NOT "v(dv/dx)" and, by the chain rule,
$$= \frac{dy}{dx}\frac{dv/dy}= v\frac{dv}{dy}$$

That's a method called "quadrature" and is often used where the independent variable does not appear explicitely in the equation.

Unfortunately, once you have solved for v as a function of y, the integral for y as a function of x tends to be wicked! Often leading to an "elliptic integral" which I suspect is the case here.

 

[gabbagabbahey]

Oops, yes, I think I need some coffee

 

[HallsofIvy]

By the way, here's an interesting application of "quadrature". Suppose you have a particle moving, on the real number line, under the influence of a force that depends only on position x and not t.

The equation of motion, "force= mass times acceleration", is
$$m \frac{d^2 x}{dt^2}= f(x)$$
where f(x) is the force function. Since t does not appear explicitely (the equation is "autonomous") we can let v= dx/dt and, as before, the equation becomes
$$mv\frac{dv}{dt}= f(x)$$
a separable equation. Write it as mv dv= f(x)dx and integrate both sides:
$$\frac{1}{2}mv^2= \int f(x)dx+ C$$
or
$$\frac{1}{2}mv^2- \int f(x)dx= C$$

Do you see that that is "conservation of energy"? $(1/2)mv^2$ is the kinetic energy and $-\int f(x)dx$ is the potential energy.

 

[mst_ab]

?
y'=v=>(dy/dx)=v *
y''=dv/dx=(dv/dx)(dy/dx)=v(dv/dy) =_*_=> y"=v(dv/dy)

 

[HallsofIvy]

Oops, yes, I think I need some coffee

Hey, he said he was a University Professor and Universtiy Professors NEVER make mistakes!

 

[gabbagabbahey]

Hey, he said he was a University Professor and Universtiy Professors NEVER make mistakes!

Of course not, but apparently they do get "homework"