More Differential Equations....

[MermaidWonders]

Question - True or False: If $\frac{dx}{dt}$ = $\frac{1}{x}$ and $x$ = 3 when $t$ = 0, then $x$ is an increasing function of $t$.

View attachment 8011

I understand how the graph of $x$ was obtained (the graph on the board), but I really don't understand why she attempted to draw the negative root of $x$ the way she did in those faint dash lines (if you can see them, they actually try to show a the function $-x$). In other words, why is the $-x$ function a reflection in the $y$-axis of function $x$? Why isn't it reflected in the $x$-axis?

 

[HOI]

Question - True or False: If $\frac{dx}{dt}$ = $\frac{1}{x}$ and $x$ = 3 when $t$ = 0, then $x$ is an increasing function of $t$.
I understand how the graph of $x$ was obtained (the graph on the board), but I really don't understand why she attempted to draw the negative root of $x$ the way she did in those faint dash lines (if you can see them, they actually try to show a the function $-x$). In other words, why is the $-x$ function a reflection in the $y$-axis of function $x$? Why isn't it reflected in the $x$-axis?

Yes, if dx/dt= 1/x then x dx= dt. Integrating both sides, (1/2)x^2= t+C. Since x= 3 when t= 0, 9/2= C. (1/2)x^2= t+ 9/2 or x^2= 2t+ 9. Solving for x= sqrt(2t+ 9).

Strictly speaking, there is no "y" in the problem so there is no y-axis! You are really asking "why is only the positive square root taken and not x= -sqrt(2t+ 9)?" The difficulty is that 1/x does not exist for x= 0 so dx/dt= 1/x does not exist at x= 0. The initial value, x= 3 when t= 0 is in the upper half plane and the solution cannot be extended past x= 0.

 

[MarkFL]

I believe the faint graph is the imaginary part of the solution. I don't know why it would be included though.

Personally, I wouldn't even compute the solution to answer the question...given that $x$ is positive at $t=0$, and the fact that the RHS of the ODE is not going to change sign, we must therefore have that $x$ is an increasing function of $t$.

 

[MermaidWonders]

Yes, if dx/dt= 1/x then x dx= dt. Integrating both sides, (1/2)x^2= t+C. Since x= 3 when t= 0, 9/2= C. (1/2)x^2= t+ 9/2 or x^2= 2t+ 9. Solving for x= sqrt(2t+ 9).

Strictly speaking, there is no "y" in the problem so there is no y-axis! You are really asking "why is only the positive square root taken and not x= -sqrt(2t+ 9)?" The difficulty is that 1/x does not exist for x= 0 so dx/dt= 1/x does not exist at x= 0. The initial value, x= 3 when t= 0 is in the upper half plane and the solution cannot be extended past x= 0.

Wait... so does $x$ depend on $t$? Then why doesn't the graph show the point (0, 3) due to the fact that $x$ = 3 and $t$ = 0? Put another way, why does the graph pass through the origin instead?

- - - Updated - - -

I believe the faint graph is the imaginary part of the solution. I don't know why it would be included though.

Personally, I wouldn't even compute the solution to answer the question...given that $x$ is positive at $t=0$, and the fact that the RHS of the ODE is not going to change sign, we must therefore have that $x$ is an increasing function of $t$.

How do you know that "the RHS of the ODE is not going to change sign" everywhere else?

 

[skeeter]

$\dfrac{dx}{dt} = \dfrac{1}{x} \implies x = f(t)$

solving the DE with the given initial condition $(0,3)$ yields ...

$x(t) = \sqrt{2t+9}$

$\dfrac{dx}{dt} = \dfrac{1}{\sqrt{2t+9}} = \dfrac{1}{x}$

 

[MermaidWonders]

$\dfrac{dx}{dt} = \dfrac{1}{x} \implies x = f(t)$

solving the DE with the given initial condition $(0,3)$ yields ...

$x(t) = \sqrt{2t+9}$

$\dfrac{dx}{dt} = \dfrac{1}{\sqrt{2t+9}} = \dfrac{1}{x}$

Wait... so why didn't my prof draw the graph like the one you have here (which actually passes through (0, 3))? Is it because her graph is the general solution without taking into account the initial conditions?

 

[MermaidWonders]

Now, I'm still confused why my professor drew the very faint dashed graph that is the reflection of the "main" graph there... :(

 

[skeeter]

Wait... so why didn't my prof draw the graph like the one you have here (which actually passes through (0, 3))? Is it because her graph is the general solution without taking into account the initial conditions?

I have no idea.

The chalkboard graph is when $C = 0$, $x = \sqrt{2t}$

the reflected graph is $x = \sqrt{2(-t)}$ ... I'd ask your prof for further clarification

 

[MermaidWonders]

I have no idea.

The chalkboard graph is when $C = 0$, $x = \sqrt{2t}$

the reflected graph is $x = \sqrt{2(-t)}$ ... I'd ask your prof for further clarification

OK, I'll do that. Thanks!