# Very simple differential equation

• May 18th 2008, 03:30 PM
erorr
Very simple differential equation
dy/dt = t^3 * y^3

This is supposed to be solved using seperation of variables, but after seperating and integrating both sides, you end up with y^-2 on one side and a negative number on the other side (after you multiply both sides by -2 to begin to solve for y). Solving for y would require rasing each side to the -1/2 power... but you can't take the square root of a negative number. That is where I'm stuck.

edit: here is what I end up with:

y^-2 = -1/2 * t^4
• May 18th 2008, 03:59 PM
Mathstud28
Quote:

Originally Posted by erorr
dy/dt = t^3 * y^3

This is supposed to be solved using seperation of variables, but after seperating and integrating both sides, you end up with y^-2 on one side and a negative number on the other side (after you multiply both sides by -2 to begin to solve for y). Solving for y would require rasing each side to the -1/2 power... but you can't take the square root of a negative number. That is where I'm stuck.

edit: here is what I end up with:

y^-2 = -1/2 * t^4

This is an SDE

So we begin by seperating

$\displaystyle \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y ^3}=t^3dt}$

Integrating we get

$\displaystyle \frac{-1}{2y^2}=\frac{t^4}{4}+C$

Solving we get

$\displaystyle \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}$

I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?
• May 18th 2008, 04:56 PM
Krizalid
Quote:

Originally Posted by Mathstud28

I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?

The thing is that you solved wrong the following:

Quote:

Originally Posted by Mathstud28

Integrating we get

$\displaystyle \frac{-1}{2y^2}=\frac{t^4}{4}+C$

Solving we get

$\displaystyle \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}$

• May 18th 2008, 05:07 PM
TheEmptySet
Quote:

Originally Posted by Mathstud28
This is an SDE

So we begin by seperating

$\displaystyle \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y ^3}=t^3dt}$

Integrating we get

$\displaystyle \frac{-1}{2y^2}=\frac{t^4}{4}+C$

Solving we get

$\displaystyle \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}$

I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?

The C got lost(Thinking)

$\displaystyle \frac{-1}{2y^2}=\frac{t^4}{4}+C$

$\displaystyle \frac{-1}{2y^2}=\frac{t^4+C_1}{4}$

$\displaystyle y^2=\frac{4}{-2t^4+C_2}$

$\displaystyle y=\pm \frac{2}{\sqrt{-2t^4+K}}$

I hope this helps

Good luck.
• May 18th 2008, 05:13 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
This is an SDE

So we begin by seperating

$\displaystyle \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y ^3}=t^3dt}$

Integrating we get

$\displaystyle \frac{-1}{2y^2}=\frac{t^4}{4}+\color{red}C$

Solving we get

$\displaystyle \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}$

I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?

Don't forget about the C! That will change the answer a bit...it may get rid of the pesky $\displaystyle i$ term.

$\displaystyle y^{-2}=C-\frac{t^4}{2} \implies y=\frac{1}{\sqrt{C-\frac{1}{2}t^4}}$