Help with Differential Equation Problem

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Help with Differential Equation Problem!

Homework Statement

The problem is shown as:
Consider the differential equation dy/dt = (y)^(1/4), y>=0.
(a) Find all points in the (t,y) plane through which at least one solution passes. Give reasons for your answer.
(b) Prove that there is a unique solution to the differential equation that satisfies y(0) = 3. Note that you should not try to find the solution to answer this part of the question
(c) Find 2 solutions to the IVP consisting of the differential equation above with the initial condition y(1)=0.

Homework Equations

The Attempt at a Solution

I have only got this solution and have no idea about what to do with part (a) and (b) of this question.
dy/dt = (y)^(1/4)
(y)^(-1/4) dy/dt = 1
(y)^(-1/4) dy = 1 dt
int (y)^(-1/4) dy = int 1 dt
(4/3)(y)^(3/4) = t + C
(y)^(3/4) = (3/4)t + C
y = [ (3/4)t + C ]^(4/3)

Using the intial condition y(1) = 0 I got the c value as -(3/4). So I've only got one solution to this prob in the form y = [(3/4)t - (3/4)]^(4/3) and I can't find anything similar to part (a) and (b) anywhere in the text. HELP PLEASE!

 

[mighty2000]

Dear student,

I understand that this problem may seem challenging at first, but let's break it down and tackle it step by step.

(a) To find all points in the (t,y) plane through which at least one solution passes, we need to consider the initial condition y(0) = 3. This means that at t = 0, the value of y is 3. Now, we can use the equation we found in your attempt at a solution, y = [(3/4)t + C]^(4/3), to find the values of t that satisfy this initial condition. Plugging in y = 3 and t = 0, we get:

3 = [(3/4)(0) + C]^(4/3)

3 = [C]^(4/3)

C = 3^(3/4)

Therefore, the point (0,3^(3/4)) is one point in the (t,y) plane through which at least one solution passes. Can you think of any other points? Keep in mind that the value of C can be any real number.

(b) To prove that there is a unique solution to the differential equation that satisfies y(0) = 3, we can use the existence and uniqueness theorem for initial value problems. This theorem states that if a differential equation is continuous in both t and y and satisfies a Lipschitz condition (meaning there is a constant K such that |f(t,y1) - f(t,y2)| <= K|y1 - y2| for all t and y1, y2), then there exists a unique solution for any given initial condition.

In this case, our differential equation is continuous and satisfies the Lipschitz condition (you can verify this by taking the derivative of f(t,y) = (y)^(1/4) with respect to y and showing that it is bounded by a constant). Therefore, by the existence and uniqueness theorem, there is a unique solution to the differential equation that satisfies y(0) = 3.

(c) To find 2 solutions to the IVP consisting of the differential equation with the initial condition y(1) = 0, we can use the same method as in part (a). Plugging in y = 0 and t = 1 into the equation y = [(3/4)t + C]^(4/3), we get:

0 = [(3/4