Question

Form the differential eq. for the function y=a*cosh(x/a) a= arbituary constant

Answer 1

Your question doesn't make much since, can you elaborate?

Answer 2

i think yahya means to find the diff equation that its answer is y=a*cosh(x/a)

Answer 3

eliminate arbituary constant 'a' and form the differential equation for the given function.

I tried my best

Answer 4

the differential equation will be of first order

Answer 5

Means u have to take the derivative in order to get a differential equation..... Just derivate this..... U will get the answer........

Answer 6

ok let me do some thinking here\[y=a \cosh(\frac{x}{a})\]\[y'= \sinh(\frac{x}{a})\]\[(\frac{y}{a})^2-(y')^2=\cosh^2(\frac{x}{a})-\sinh^2(\frac{x}{a})=1\]

Answer 7

I solve this like that..... Can anybody guide me if I am wrong..

Answer 8

u'r not wrong :) but thats not the requierd thing...

Answer 9

dy/dx=sinh(x/a) and now if u will take the integration as we solve in differential equations...... U will get the above result..... then what should I do further ?

Answer 10

\[(\frac{y}{a})^2-(y')^2=1\]\[y'^2=\frac{y^2}{a^2}-1\]first order but nonlinear ordinary differential equation

Answer 11

I dont know from where u write it (y/a)^2-..... the whole term

Answer 12

still a is not eliminated...i tried to diff. again, and got y.y"-y'^2 =1
but don't know whether 2nd diff. is allowed...
u have options or answer key ?

Answer 13

\[ y=a\cosh\left(\frac xa\right)\]
\[ \frac{\text dy}{\text dx}=\sinh\left(\frac xa\right)\]

Answer 14

differentiate only once because it must be a first order differential equation without any arbituary constant

Answer 15

Its answer is
y*ln(1+y'2+y')=x(1+y'2)'0.5

but i can't solve it, an unsolved problem