1. ## ODE finding constants...

another problem i worked half way on and stuck on

ODE is dx/dt = x^(3/4)

i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

ok, so here goes my crazy method in doing this..

x'(t) = ba(at)^(b-1)

put this into the ODE:

ba(at)^(b-1) = ((at)^b)^3/4

now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation

thank you

2. Originally Posted by matlabnoob
another problem i worked half way on and stuck on

ODE is dx/dt = x^(3/4)

i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

ok, so here goes my crazy method in doing this..

x'(t) = ba(at)^(b-1)

put this into the ODE:

ba(at)^(b-1) = ((at)^b)^3/4

now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation

thank you
$\frac{dx}{dt}=x^{3/4}\implies x^{-3/4}dx=dt\implies \int x^{-3/4}\,dx=\int\,dt$ $\implies 4x^{1/4}=t+c\implies x=\left(\frac{1}{4}(t+c)\right)^4$

(with $c$ being an arbitrary constant). So to model your equation to match the one given, just let $c=0$, and then it follows that $a=\frac{1}{4}$ and $b=4$.

3. Originally Posted by matlabnoob
another problem i worked half way on and stuck on

ODE is dx/dt = x^(3/4)

i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

ok, so here goes my crazy method in doing this..

x'(t) = ba(at)^(b-1)

put this into the ODE:

ba(at)^(b-1) = ((at)^b)^3/4

now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation

thank you
$ba(at)^{b-1}= ba(a^{b-1})t^{b-1}= ba^b t^{b-1}$ and $((at)^b)^{3/4}= (at)^{3b/4}= a^{3b/4}t^{3b/4}$.

In order that those be equal for all t, we must have $ba^b= a^{3b/4}$ and b- 1= 3b/4.

That last equation gives 4b- 4= 3b so b= 4. With that, the first equation becomes $4a^4= a^3$ so either a= 0 or a= 1/4.

4. Originally Posted by redsoxfan325
$\frac{dx}{dt}=x^{3/4}\implies x^{-3/4}dx=dt\implies \int x^{-3/4}\,dx=\int\,dt$ $\implies 4x^{1/4}=t+c\implies x=\left(\frac{1}{4}(t+c)\right)^4$

(with $c$ being an arbitrary constant). So to model your equation to match the one given, just let $c=0$, and then it follows that $a=\frac{1}{4}$ and $b=4$.

thankks! ok..so what happens once i find a solution to this ODE... and then im told to...

'find another solution of the ODE,different from the one i just found, which exists for all t in R (real numbers)'

any hints, steps...anything! would be soo appreciated