Solving differential equations: Help explain the answers please

**ironz** · December 22nd 2011, 06:30 AM

I have done most of the questions, I am just struggling on a few parts. The question is:

a) Use the method of Separation of Variables to find the General Solution.
b) Analyse the right hand side and determine for what pairs $(x_0,y_0)$ the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

1) $\frac{dy}{dx} = y^{\frac{1}{2}}$ . IC = y(1) = 0.

Answers:

a) I got: $y = (\frac{2}{3}(x + C) )^{\frac{3}{2}}$ , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where $G(y) = y^{\frac{3}{2}} = 0 \implies y = 0$ . My first question is: What does the special solution do?

b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

The solutions to the IVP exist for any pair $(x_0,y_0) \in R$ (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of $x_0$ , but not when $y_0 = 0$ and so the interval would be $(x_0,y_0) \in (-\infty,0) \cup (0,\infty)$ , again, they however multiplied both intervals by some 'R'.

I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.

c) This is the main one I am having trouble with. This is what it says in the answers:

The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have $0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}}$ , hence C = 0 and $y = (\frac{2x}{3})^{\frac{3}{2}}$ . This general form is a "compound" solution (Whatever that means)of the form

y = { $0 , x \in (\infty, B)$ or $(\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)$

depending on one arbitary parameter $B \geq 0$ . Each of these solutions is defined for all $I = (-\infty, \infty)$ , and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.

Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

Thank you

**chisigma** · December 24th 2011, 01:28 AM

Originally Posted by ironz

I have done most of the questions, I am just struggling on a few parts. The question is:

a) Use the method of Separation of Variables to find the General Solution.
b) Analyse the right hand side and determine for what pairs $(x_0,y_0)$ the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

1) $\frac{dy}{dx} = y^{\frac{1}{2}}$ . IC = y(1) = 0.

Answers:

a) I got: $y = (\frac{2}{3}(x + C) )^{\frac{3}{2}}$ , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where $G(y) = y^{\frac{3}{2}} = 0 \implies y = 0$ . My first question is: What does the special solution do?

b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

The solutions to the IVP exist for any pair $(x_0,y_0) \in R$ (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of $x_0$ , but not when $y_0 = 0$ and so the interval would be $(x_0,y_0) \in (-\infty,0) \cup (0,\infty)$ , again, they however multiplied both intervals by some 'R'.

I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.

c) This is the main one I am having trouble with. This is what it says in the answers:

The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have $0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}}$ , hence C = 0 and $y = (\frac{2x}{3})^{\frac{3}{2}}$ . This general form is a "compound" solution (Whatever that means)of the form

y = { $0 , x \in (\infty, B)$ or $(\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)$

depending on one arbitary parameter $B \geq 0$ . Each of these solutions is defined for all $I = (-\infty, \infty)$ , and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.

Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

Thank you

Looking at the DE...

$y^{'}=\sqrt{y}$ (1)

... You observe that, because the x is not in the DE, if $\varphi(x)$ is solution of (1), then $\varphi(x-\xi)$ with $\xi$ arbitrary is also solution of (1). By simple inspection You find that...

$\varphi(x)=\begin{cases}\frac{x^{2}}{4} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0\end{cases}$ (2)

... is solution of (1). Now the 'initial condition' is $y(1)=0$ and that means that any $\varphi(x-\xi)$ with $\xi \ge 1$ is solution, so that it doesn't exist only one solution...

Marry Christmas from Serbia

$\chi$ $\sigma$

**HallsofIvy** · December 27th 2011, 06:49 AM

Originally Posted by ironz

I have done most of the questions, I am just struggling on a few parts. The question is:

a) Use the method of Separation of Variables to find the General Solution.
b) Analyse the right hand side and determine for what pairs $(x_0,y_0)$ the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

1) $\frac{dy}{dx} = y^{\frac{1}{2}}$ . IC = y(1) = 0.

Answers:

a) I got: $y = (\frac{2}{3}(x + C) )^{\frac{3}{2}}$ , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where $G(y) = y^{\frac{3}{2}} = 0 \implies y = 0$ . My first question is: What does the special solution do?

b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

The solutions to the IVP exist for any pair $(x_0,y_0) \in R$ (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of $x_0$ , but not when $y_0 = 0$ and so the interval would be $(x_0,y_0) \in (-\infty,0) \cup (0,\infty)$ , again, they however multiplied both intervals by some 'R'.

I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.

$R\times R$ or $R^2$ is the "Cartesian product"- the set of ordered pairs of real numbers. Since each of $x_0$ and $y_0$ are real numbers, the pair $(x_0, y_0)$ is in $R\times R$ , not in R which contains numbers, not pairs of numbers.

c) This is the main one I am having trouble with. This is what it says in the answers:

The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have $0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}}$ , hence C = 0 and $y = (\frac{2x}{3})^{\frac{3}{2}}$ . This general form is a "compound" solution (Whatever that means)of the form

y = { $0 , x \in (\infty, B)$ or $(\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)$

depending on one arbitary parameter $B \geq 0$ . Each of these solutions is defined for all $I = (-\infty, \infty)$ , and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.

Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

Thank you

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