1) dx/dt=sgn(x), x(0) =0;  2) dx/dt=sgn(t), x(0)=0; 3) dx/dt=3*x^(2/3),  x(0) =0; 4) dx/dt=1+x^2  t={-pi/2 pi/2} Find solutions of the form x = f(t) where (f is a one-one function) in each...

1) dx/dt=sgn(x), x(0) =0; 

2) dx/dt=sgn(t), x(0)=0;

3) dx/dt=3*x^(2/3),  x(0) =0;

4) dx/dt=1+x^2  t={-pi/2 pi/2}

Find solutions of the form x = f(t) where (f is a one-one function) in each case if any exist and state whether there is a unique solution.

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mathsworkmusic | (Level 2) Educator

Posted on

The question is asking for solutions to the given differential equations in each case, where the initial value is given (that is, they are Initial Value Problems). If a solution does indeed exist we are asked whether it is unique or not.

1) `dx/dt = sgn(x)`  where `x(0) = 0`

'sgn' is the sign or signum function, which is +1 when the variable is positive and -1 when the variable is negative.

So here when x<0 we have that  `dx/dt = -1` ````and when x>0 we have that `dx/dt = 1`

The initial value given tells us that when x=0, t=0.

So the solution x = f(t) passes through the origin (x,t) = (0,0) and has gradient +1 when x is positive and gradient -1 when x is negative. The solution is then either

x = t  or  x = -t   and is valid only on the range t `in [0,infty]`

Therefore two solutions exist, neither of which then are unique (the solution looks like a V-shape, symmetric about the positive t axis).

2) `dx/dt = sgn(t)`  where `x(0) = 0`

Here, when t<0 we have that

`dx/dt = -1` and when t>0 we have that `dx/dt = 1`

``The initial value given tells us that when x=0, t=0.

As in 1), the solution x = f(t) passes through the origin (x,t) = (0,0) , but now has gradient +1 when t (rather than x) is positive and gradient -1 when t (rather than x) is negative. The solution is in this case unique and described by the graph

x = |t|

which is a V-shape again, but this time symmetric about the positive x axis, giving a valid one-one function of t.

3) `dx/dt = 3x^(2/3)` `t in [-pi/2,pi/2]`where x(0) = 0

To solve this differential equation, move all terms in x to the lefthand side of the equation and all terms in t to the righthand side, giving

`dx/(x^(2/3)) = 3dt`

Then integrate both sides, giving

`3x^(1/3) = 3t + c`  where `c` is a constant of integration.

From the initial value given, x(0) = 0, so that we have c=0 and hence the solution

`x^(1/3) = t`  that is   `x = t^3`

4) `dx/dt = 1 + x^2`   where `t in [-pi/2, pi/2]`

Again, move all terms in x to the lefthand side and all terms in t to the righthand side giving

`dx/(1+x^2) = dt`

Integrating on both sides gives the solution

`tan^(-1)(x) = t + c`  where again c is a constant of integration.

No initial value is given so all curves of the form

x = tan(t +c)  `t in [-pi/2,pi/2]` are valid solutions and hence not unique

Sources:
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alimudheher | eNotes Newbie

Posted on

thanx very much for ur answer 

can u show me step by step differential equation for first and second ?

 and for fourth one the x(0)=0 . 

and can u explain the uniqueness and existence in another way for each one of them 

and thank you again 

ali

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