# Math Help - Ivp!!!???

1. ## Ivp!!!???

Question:

"y1 (t) = 1/√(1-2t) and y2 (t) = 1/√(4-2t) are solutions of dy/dt = y^3. Use the plot of the solutions to answer the following parts of this question: (a) consider the IVP consisting of the differential equation given together with the initial condition y(0)=1. Find the interval of t for which the solution to the IVP exists. (b) What can you say about the behaviour of a solution to this differential equation which has an initial condition satisfying 0.5<y(0)<1.0?"

I've plotted these solutions and uploaded a picture of the graph. Help will be appreciated...thank you

2. Originally Posted by Ghostrider
Question:

"y1 (t) = 1/√(1-2t) and y2 (t) = 1/√(4-2t) are solutions of dy/dt = y^3. Use the plot of the solutions to answer the following parts of this question: (a) consider the IVP consisting of the differential equation given together with the initial condition y(0)=1. Find the interval of t for which the solution to the IVP exists. (b) What can you say about the behaviour of a solution to this differential equation which has an initial condition satisfying 0.5<y(0)<1.0?"

I've plotted these solutions and uploaded a picture of the graph. Help will be appreciated...thank you
(a) y1 (t) = 1/√(1-2t) is the solution corresponding to y(0) = 1. Solutions exists for 1 - 2t > 0 => t < 1/2. You can also see this from the graph.

(b) y2 (t) = 1/√(4-2t) is the solution corresponding to y(0) = 0.5.

I'd discuss domain, shape and behaviour as t --> -oo.

Note that the general solution is $y = \pm \frac{1}{\sqrt{A - 2t}}$ and $y(0) = \pm \frac{1}{\sqrt{A}}$. Domain: $t < \frac{A}{2}$.

$\frac{1}{2} < \frac{1}{\sqrt{A}} < 1 \Rightarrow 1 < A < 4$.