Question

\frac{\frac{d^{2}v}{dx^{2}}}{[1+(\frac{dv}{dx})^{2}]^{\frac{3}{2}}}=cM(x)
Find v=?

Answer 1

What is this problem, physically?

Answer 2

\[\frac{\frac{d^{2}v}{dx^{2}}}{[1+(\frac{dv}{dx})^{2}]^{\frac{3}{2}}}=cM(x)\]

Answer 3

It's equation of deflection of beam

Answer 4

Oh wait. Write w = dv/dx. Then this equation is

w' / (1 + w^2)^(3/2) = = c M(x)

Answer 5

and this equation is separable:
\[\int\limits \frac{dw}{ (1+w^2)^{3/2}} \ = \ \int\limits cM(x) dx\]

Answer 6

Now integrate, solve for w, then for v.

Answer 7

after integrating
\[\frac{v'}{\sqrt{1+(v')^{2}}+C_1=c\int M(x)dx\]

Answer 8

\[\frac{v'}{\sqrt{1+(v')^{2}}}+C_1=c\int M(x)dx\]
still not solved

Answer 9

But once you have, you'll find the expression in w is quite messy and you'll be all but convinced that it's very difficult to write down a general solution for v.

Which is why you need to go back to the problem; try and figure out more about v and the function M(x).

Answer 10

Oh, sorry, yes you have written it down correctly. My mistake. But even so, solving for v' even now is still not pretty at all.

Answer 11

put w=tan(@)
[\dw=sec^{2}(@) d@\]
[\int cos(@) d@\]
so =sin@
=w/sqrt(1+w^2) +C1
since w=v'
=v'/sqrt(1+(v')^2) +C1

Answer 12

I..e, you can go through the algebra and find an expression for v' and then say v is the integral of that expression.

But it's not going to be simple.

Answer 13

I know it's a non linear equation and it's not simple. But I just want to find out it's solution.

Answer 14

I.e.,
\[\frac{ v'^2 }{1 +v'^2} = ( c \int\limits M(x) dx - c_1)^2\]
and now solve for v'.

Answer 15

Well, keep going then. You can write down the expression.

Answer 16

If I suppose v as power series then obviously the problem will not be solved

Answer 17

No, I don't see how with a general M(x) power series help you.

Answer 18

I found one
Let \[f(x)=c\int M(x) dx-C_1\]

\[v'^{2}=(1+(v')^{2})f(x)^{2}\]
\[v'^{2}(1-f(x)^{2})=f(x)^{2}\]
\[v'^{2}=\frac{f(x)^{2}}{1-f(x)^{2}}\]
\[v'=\sqrt{\frac{f(x)^{2}}{1-f(x)^{2}}}\]
so
\[v=\int\sqrt{\frac{f(x)^{2}}{1-f(x)^{2}}}dx+C_2\]

Answer 19

Thanks for your sincere help.