Question

use separation of variables to solve the initial value problem dy/dt= y^2 and observe that the solution has a vertical asymptote at t=1. this phenomenon is called blow-up..because the solution becomes infinite in a finite time. but then it also asks to solve the initial problem using modified euler's method in excel

Answer 1

where is your initial value?

Answer 2

In general, if you have
\(\color{#000000}{ \displaystyle \frac{dy}{dt} =y^n }\)

You can bring the y-terms to the other side by division.
\(\color{#000000}{ \displaystyle y^{-n}\frac{dy}{dt} =1 }\)

then integrate both sides with respect to t.
\(\color{#000000}{ \displaystyle \int y^{-n}\frac{dy}{dt} dt= \int 1 dt }\)

Note that the dt cancels on the left side.
\(\color{#000000}{ \displaystyle \int y^{-n}dy= \int dt }\)

Then, for \(n\ne 1\), you get:

\(\color{#000000}{ \displaystyle \frac{y^{-n+1}}{-n+1}=x+C }\)

\(\color{#000000}{ \displaystyle y^{-n+1}=(-n+1)x+(-n+1)C }\)

Number (-n+1) × arbitrary constant C is still arbitrary constant C, so,

\(\color{#000000}{ \displaystyle y^{-n+1}=(-n+1)x+C }\)

and you can write it explicitely if -n+1 is odd.

Answer 3

then, if you are given that y(a)=b, then

\(\color{#000000}{ \displaystyle a^{-n+1}=(-n+1)b+C }\)
\(\color{#000000}{ \displaystyle a^{-n+1}-(-n+1)b=C }\)
\(\color{#000000}{ \displaystyle a^{-n+1}+(n-1)b=C }\)

So,
\(\color{#000000}{ \displaystyle y^{-n+1}=(-n+1)x+a^{-n+1}+(n-1)b }\)

Answer 4

but, this is all abstract of course.
You would get a simpler result if you tried the same process with your equation.