Looking for an explanation of a simple Lipshitz condition

[dynamicskillingme]

Homework Statement

Has solution

It then goes on to state the solution blows up at

, which I understand.

My issue is when I do the solution I get

(Working)

 

[Stephen Tashi]

$\dot{y} = y^2$
$y = yt + c$ ; for $t = 0$

You seem to be treating $y$ as if it were a constant instead of a function of $t$.
We can't assume $\int y dt = (y)(t) = yt$ because, in this problem the antiderivative needed is $\int y(t) dt$ where $y(t)$ is an unknown function.

For example, if $y(t) = e^t$ then $\int y(t) dt = e^t + c \ne (y(t))(t)$.

 

[dynamicskillingme]

If you could humour me as an engineer having to take a maths course and is struggling to remember concepts taught along time ago, How could I prove their answer, or what should I read up on to be able to show this proof?

 

[Ray Vickson]

Homework Statement

Has solution

It then goes on to state the solution blows up at

, which I understand.

My issue is when I do the solution I get

(Working)

Wrong: $y$ is a function of $t$ because, after all, you have $\dot{y} = dy/dt$ in your equation to start with! So, from $dy/dt = y^2$ you have
$$\frac{dy}{y^2} = dt \: \Longrightarrow -y^{-1} = t + c,$$
or
$$\frac{1}{y} = -c-t \: \Longrightarrow y = \frac{-1}{t+c}$$
Set $c = -1/y_0$ and see what you get.

Anyway, if all you want to do is to see why $y = y(t) = y_0/(1-y_0 t)$ is a solution, just plug it into your differential equation $dy/dt = y^2$ and see if it works.

 

[Stephen Tashi]

How could I prove their answer, or what should I read up on to be able to show this proof?

Does the problem simply want to you show that $y(t) = \frac{y0}{1 - y0 t}$ is a solution ? You do that by substituting that function in for $y$ in the differential equation. (Of course you also have to substitute the derivative of $\frac{y0}{1 - y0 t}$ in for $\dot{y}$ , which means remembering enough calculus to compute that derivative ).When you substitute a solution into a differential equation, the equation should become an identity - i.e. the two sides should be the same function when you simplify things.Solving differential equations when you aren't given a solution, often involves guessing it. If you have to find the solution by yourself, start by reviewing your knowledge of derivatives with emphasis on noting how derivatives of a function y can be expressed in terms of the original function y.

For example:

$\dot{y} = 5y$ corresponds to the derivative of $e^{5t}$ is $5 e^{5t}$.

Using "$D$" to denote the operation of taking a derivative, $De^{kt} = k e^{kt}$. So when you are asked to solve a differential equation $\dot{y} = k y$ (where $k$ is a constant) you should be able to guess the answer is an exponential function.Ray Vickson illustrated the technique of solving a "separable" differential equation. You can find videos on the web that explain that technique in detail.