# Thread: Constant(?) Solution of a Separable ODE.

1. ## Constant(?) Solution of a Separable ODE.

I'm not sure if I understand this question. By separating variables and integrating, $\displaystyle y(x) = x^{3/2}$ is an obvious solution.$\displaystyle y(x) = 0$ seems to imply that $\displaystyle y(x) = x^{3/2} + c = 0$, so $\displaystyle x^{3/2} = -c$ which implies that c is not constant...? Am I missing something?

This is problem 4-a in the following assignment:
http://ocw.mit.edu/NR/rdonlyres/Math...s09_pset01.pdf

The corresponding solution set can be found:
http://ocw.mit.edu/NR/rdonlyres/Math...sol_pset01.pdf

I do not understand their solution at all.

Thanks in advance.

2. Originally Posted by davismj

I'm not sure if I understand this question. By separating variables and integrating, $\displaystyle y(x) = x^{3/2}$ is an obvious solution.$\displaystyle y(x) = 0$ seems to imply that $\displaystyle y(x) = x^{3/2} + c = 0$, so $\displaystyle x^{3/2} = -c$ which implies that c is not constant...? Am I missing something?

This is problem 4-a in the following assignment:
http://ocw.mit.edu/NR/rdonlyres/Math...s09_pset01.pdf

The corresponding solution set can be found:
http://ocw.mit.edu/NR/rdonlyres/Math...sol_pset01.pdf

I do not understand their solution at all.

Thanks in advance.
Dear davismj,

Consider the original equation, $\displaystyle \frac{dy}{dx}=\frac{3}{2}y^{\frac{1}{3}}$

Here y(x)=0 is a obvious solution. Since if y(x)=0, RHS=0 and LHS=0. Hence the equation is satisfied. However when you divide the equation by $\displaystyle y^{\frac{1}{3}}$ to obtain, $\displaystyle dx=\frac{2}{3}y^{-\frac{1}{3}}dy$ you make the assumption that $\displaystyle y\neq{0}$, therefore you cannot argue that "....imply that $\displaystyle y(x) = x^{3/2} + c = 0$ so $\displaystyle x^{3/2} = -c$ which implies that c is not constant...."

Hope this will help you to undestand.

3. Originally Posted by Sudharaka
Dear davismj,

Consider the original equation, $\displaystyle \frac{dy}{dx}=\frac{3}{2}y^{\frac{1}{3}}$

Here y(x)=0 is a obvious solution. Since if y(x)=0, RHS=0 and LHS=0. Hence the equation is satisfied. However when you divide the equation by $\displaystyle y^{\frac{1}{3}}$ to obtain, $\displaystyle dx=\frac{2}{3}y^{-\frac{1}{3}}dy$ you make the assumption that $\displaystyle y\neq{0}$, therefore you cannot argue that "....imply that $\displaystyle y(x) = x^{3/2} + c = 0$ so $\displaystyle x^{3/2} = -c$ which implies that c is not constant...."

Hope this will help you to undestand.
Ahh, I understand. Thank you.

4. Notice that you are [b]not[/b ]asked to solve this equation!

You are, instead, given two functions and asked to show that they are solutions. That is much easier. (Think about the difference between solving the equation [/tex]xe^{x-1}+ x^2- 2= 0[/tex] and showing that x= 1 is a solution!)

If y= 0, then, of course, y'= 0 so the equation, y'= (3/2)y^{1/3} becomes "0= 0".

If $\displaystyle y= x^{3/2}$, then $\displaystyle y'= (3/2)x^{3/2- 1}= (3/2)x^{1/2}= (2/3)y^{1/3}$ because $\displaystyle y= x^{3/2}$ gives immediately that $\displaystyle y^{1/3}= x^{1/2}$

5. Originally Posted by HallsofIvy
Notice that you are [b]not[/b ]asked to solve this equation!

You are, instead, given two functions and asked to show that they are solutions. That is much easier. (Think about the difference between solving the equation [/tex]xe^{x-1}+ x^2- 2= 0[/tex] and showing that x= 1 is a solution!)

If y= 0, then, of course, y'= 0 so the equation, y'= (3/2)y^{1/3} becomes "0= 0".

If $\displaystyle y= x^{3/2}$, then $\displaystyle y'= (3/2)x^{3/2- 1}= (3/2)x^{1/2}= (2/3)y^{1/3}$ because $\displaystyle y= x^{3/2}$ gives immediately that $\displaystyle y^{1/3}= x^{1/2}$
This makes everything perfectly clear. Thank you. The following question asks me to use the fact that these two are solutions to show something else, so now I understand the connection.