# 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.

### 2 Answers | Add Yours

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:**

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