MATH1251 Questions HELP (2 Viewers)

1008 · Aug 6, 2016

Onto ODE's now:

$\\\text{(i) Solve } \frac{\mathrm{d} x}{\mathrm{d} t} =y-2y^2\text{ by making the substitution }z=\frac{1}{y}\text{ to obtain a separable differential equation in }z\\\text{and }t$

1008 · Aug 6, 2016

InteGrand · Aug 6, 2016

1008 said:
Onto ODE's now:

$\\\text{(i) Solve } \frac{\mathrm{d} x}{\mathrm{d} t} =y-2y^2\text{ by making the substitution }z=\frac{1}{y}\text{ to obtain a separable differential equation in }z\\\text{and }t$

(For interest's sake, this is a Bernoulli differential equation: https://en.wikipedia.org/wiki/Bernoulli_differential_equation . The method for solving it is typically via a substitution like the one suggested.)

$\noindent Let $z= y^{-1}$, $y\neq 0$, so $z^{\prime} = -y^\prime y^{-2}$. If we divide the original ODE through by $y^2$, it is $y^{-2} y^\prime = y^{-1} -2$, i.e. $-z^\prime = z -2$. In other words, $z^\prime = 2 -z$. Using standard methods or inspection (this is a linear first order ODE, so is easy to solve), we find the general solution is $z(t) = 2 + C e^{-t}$. So $y(t) = \frac{1}{2+ Ce^{-t}}$.$

$\noindent The other solution is $y\equiv 0$.$

$\noindent So basically, if we have a Bernoulli DE $y^\prime = P(x)y + Q(x)y^n$, where $n$ is a real number not equal to $0$ or $1$, the strategy to get the non-trivial solutions (i.e. solutions other than $y\equiv 0$) is to \emph{divide through by} $y^n$ to get $\frac{y^\prime}{y^n} = P(x)y^{1-n} + Q(x)$. Then substitute something like $v = y^{1-n}$, so $v^\prime = (1-n)y^{-n}y^\prime$. This means the DE gets transformed into $\frac{1}{1-n}v^\prime = P(x)v + Q(x)$, which is linear and can be solved via integrating factor method. Once we get this solution for $v$, we get the solution for $y$ by using $y= v^{\frac{1}{n-1}}$.$

$\noindent In the case that $n=0$, the original DE is linear already and we'd just solve it normally (integrating factor etc.). If $n=1$, the DE is a separable one.$

1008 · Aug 6, 2016

InteGrand said:
(For interest's sake, this is a Bernoulli differential equation: https://en.wikipedia.org/wiki/Bernoulli_differential_equation . The method for solving it is typically via the substitution suggested.)

$\noindent Let $z= y^{-1}$, $y\neq 0$, so $z^{\prime} = -y^\prime y^{-2}$. If we divide the original ODE through by $y^2$, it is $y^{-2} y^\prime = y^{-1} -2$, i.e. $-z^\prime = z -2$. In other words, $z^\prime = 2 -z$. Using standard methods or inspection (this is a linear first order ODE, so is easy to solve), we find the general solution is $z(t) = 2 + C e^{-t}$. So $y(t) = \frac{1}{2+ Ce^{-t}}$.$

$\noindent The other solution is $y\equiv 0$.$

Thanks for the solution and the Wikipedia link! Any thoughts on the other question (ii)?

InteGrand · Aug 6, 2016

1008 said:
Thanks for the solution and the Wikipedia link! Any thoughts on the other question (ii)?

The method would be similar. Use the given substitution and you should end up with a linear ODE, which you can solve via an integrating factor for example. Make sure to find the value of the constant C by using the initial condition. Once you have found the solution, you should be able to find its maximum value.

(Note that maximising y will be equivalent to minimising z, and the latter will be easier to do, and y_max. = 1/z_min..)

1008 · Aug 7, 2016

InteGrand said:
The method would be similar. Use the given substitution and you should end up with a linear ODE, which you can solve via an integrating factor for example. Make sure to find the value of the constant C by using the initial condition. Once you have found the solution, you should be able to find its maximum value.

(Note that maximising y will be equivalent to minimising z, and the latter will be easier to do, and y_max. = 1/z_min..)

Thanks, figured out the ODE bit, couldn't find how to do the max. value...

Got another question as well:
$\\\text{Solve } \frac{\mathrm{d} y}{\mathrm{d} x} =(y-x)^2\text{ using the substitution }u=y-x$

I found a VERY simple solution, I don't think it's right, and it certainly doesn't match with these answers provided:

$\\2x+C=\ln\left|\frac{y-x-1}{y-x+1}\right|\text{ or }y=x-1+\frac{2}{1+De^{2x}}$

My solution was just:
$\\\frac{(y-x)^3}{3}+y=C$

InteGrand · Aug 7, 2016

1008 said:
Thanks, figured out the ODE bit, couldn't find how to do the max. value...

Got another question as well:
$\\\text{Solve } \frac{\mathrm{d} y}{\mathrm{d} x} =(y-x)^2\text{ using the substitution }u=y-x$

I found a VERY simple solution, I don't think it's right, and it certainly doesn't match with these answers provided:

$\\2x+C=\ln\left|\frac{y-x-1}{y-x+1}\right|\text{ or }y=x-1+\frac{2}{1+De^{2x}}$

My solution was just:
$\\\frac{(y-x)^3}{3}+y=C$

How did you get your solution? Note we can't just integrate the RHS of the given ODE as though we're just integrating a square, because there's an unknown function y(x) in there.

1008 · Aug 7, 2016

InteGrand said:
How did you get your solution? Note we can't just integrate the RHS of the given ODE as though we're just integrating a square, because there's an unknown function y(x) in there.

Yeah just realised I did essentially that, I used the substitution provided, rearranged it and somehow forced it into the ODE provided (and I know I wasn't supposed to do this)...But since there are two variables in the substitution u=y-x, do we need to use partial differentiation?

InteGrand · Aug 8, 2016

1008 said:
Thanks, figured out the ODE bit, couldn't find how to do the max. value...

Got another question as well:
$\\\text{Solve } \frac{\mathrm{d} y}{\mathrm{d} x} =(y-x)^2\text{ using the substitution }u=y-x$

I found a VERY simple solution, I don't think it's right, and it certainly doesn't match with these answers provided:

$\\2x+C=\ln\left|\frac{y-x-1}{y-x+1}\right|\text{ or }y=x-1+\frac{2}{1+De^{2x}}$

My solution was just:
$\\\frac{(y-x)^3}{3}+y=C$

$\noindent To obtain the solution provided by the answers, let $u= y-x$, so $y= u+x \Rightarrow y^\prime = u^\prime + 1$. Substituting into the ODE, we obtain$

$\begin{align*}u^\prime + 1 &= u^3 \\ \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} &= u^3 -1 = \left(u-1\right) \left(u^2 + u + 1\right).\end{align*}$

$\noindent Note $u\equiv 1$ is a solution (so $y= x+1$ is a solution to the original ODE. In fact, so is $y=x-1$.). Now, using separation of variables for $u\neq 1$, we have (using separation of variables)$

$\begin{align*}\int \frac{\mathrm{d}u}{\left(u-1\right)\left(u^2 + u +1\right)} &= x.\end{align*}$

$\noindent To finish from here, use partial fractions for the LHS, then remember to convert back to $y$ by replacing $u$ with $y-x$.$

InteGrand · Aug 8, 2016

1008 said:
Yeah just realised I did essentially that, I used the substitution provided, rearranged it and somehow forced it into the ODE provided (and I know I wasn't supposed to do this)...But since there are two variables in the substitution u=y-x, do we need to use partial differentiation?

$\noindent We only need regular differentiation. Recall that for these ODE's, $y$ is just a(n unknown) function of $x$, so it's a \emph{dependent} variable, so there's only one independent variable here (namely $x$), so we just differentiate normally with respect to $x$. To help make this clearer in your mind, you can write $y(x)$ and $u(x)$. So we are substituting $u(x) = y(x)-x$, so $u^\prime (x) = y^\prime (x) -1$, where $^\prime \equiv \frac{\mathrm{d}}{\mathrm{d}x}$. Once you get used to this, you can just write $u$ and $y$ instead of $u(x)$ and $y(x)$ if you want to save writing.$

1008 · Aug 8, 2016

InteGrand said:
$\noindent We only need regular differentiation. Recall that for these ODE's, $y$ is just a(n unknown) function of $x$, so it's a \emph{dependent} variable, so there's only one independent variable here (namely $x$), so we just differentiate normally with respect to $x$. To help make this clearer in your mind, you can write $y(x)$ and $u(x)$. So we are substituting $u(x) = y(x)-x$, so $u^\prime (x) = y^\prime (x) -1$, where $^\prime \equiv \frac{\mathrm{d}}{\mathrm{d}x}$. Once you get used to this, you can just write $u$ and $y$ instead of $u(x)$ and $y(x)$ if you want to save writing.$

Right, thanks. I know it's a bit off-topic, but do you mind also having a look at my other thread I just created. I think it will be mostly about first year uni probability...
http://community.boredofstudies.org/136/australian-school-business/352793/actl1101-questions-help-mostly-first-year-uni-probability.html

1008 · Aug 14, 2016

Got another ODE one...

$$\noindent A savings account is opened with a deposit of $A$ dollars. At any time $t$ years thereafter, money is being continuously deposited into the account at a rate of $(C+Dt)$ dollars per year. If interest is being paid into the account at a nominal rate of $100R\%$ per year, compounded continuously, find the balance $B(t)$ dollars in the account after $t$ years$

InteGrand · Aug 14, 2016

1008 said:
Got another ODE one...

$$\noindent A savings account is opened with a deposit of $A$ dollars. At any time $t$ years thereafter, money is being continuously deposited into the account at a rate of $(C+Dt)$ dollars per year. If interest is being paid into the account at a nominal rate of $100R\%$ per year, compounded continuously, find the balance $B(t)$ dollars in the account after $t$ years$

$\noindent By looking at what the contributions are to the instantaneous rate of increase of the balance, we see that $B^\prime (t) = C + Dt + RB(t)$, with $B(0) = A$ given. This intial value problem can be solved easily since the ODE is first-order and linear.$

1008 · Aug 20, 2016

InteGrand said:
$\noindent By looking at what the contributions are to the instantaneous rate of increase of the balance, we see that $B^\prime (t) = C + Dt + RB(t)$, with $B(0) = A$ given. This intial value problem can be solved easily since the ODE is first-order and linear.$

Thanks and sorry for the late reply.

Could someone please check my working for this one:

In the answers provided, they've got $\frac{2x^2}{3}$ instead of $\frac{2x^2}{5}$

InteGrand · Aug 20, 2016

1008 said:
Thanks and sorry for the late reply.

Could someone please check my working for this one:

In the answers provided, they've got $\frac{2x^2}{3}$ instead of $\frac{2x^2}{5}$

$\noindent You did $\exp \left(\int \frac{1}{2x}\, \mathrm{d}x\right)$ for the integrating factor, but you forgot the minus sign that is present in the DE. So the integrating factor is actually $\exp \left(\int -\frac{1}{2x}\, \mathrm{d}x\right) = \text{ reciprocal of what you got}$.$

1008 · Aug 20, 2016

InteGrand said:
$\noindent You did $\exp \left(\int \frac{1}{2x}\, \mathrm{d}x\right)$ for the integrating factor, but you forgot the minus sign that is present in the DE. So the integrating factor is actually $\exp \left(\int -\frac{1}{2x}\, \mathrm{d}x\right) = \text{ reciprocal of what you got}$.$

oooohhh, couldn't figure out my mistake yesterday. Thanks so much!

leehuan · Aug 20, 2016

I didn't have any problem tackling algebra with this question, but I have something else. (Just going to insert an image here as the question is a bit awkward to type up neatly.)

The answer to (ii) is

$y=\frac{y_0K}{e^{-kt}(K-y_0)+y_0}$

For (iii) and (v), it is quite obvious that y tends to K but I've forgotten how fractions work. Why is it that in (iii) it does so strictly increasing whereas in (iv) it is strictly increasing towards K?

(Cause I'm obviously assuming that there is obviously no need to compute dy/dt=g(t) here)

InteGrand · Aug 20, 2016

leehuan said:
I didn't have any problem tackling algebra with this question, but I have something else. (Just going to insert an image here as the question is a bit awkward to type up neatly.)

The answer to (ii) is

$y=\frac{y_0K}{e^{-kt}(K-y_0)-y_0}$

For (iii) and (v), it is quite obvious that y tends to K but I've forgotten how fractions work. Why is it that in (iii) it does so strictly increasing whereas in (iv) it is strictly increasing towards K?

(Cause I'm obviously assuming that there is obviously no need to compute dy/dt=g(t) here)

$\noindent (Note $k>0$.) If $0<y_0 < K$, then the coefficient of the decaying exponential in the denominator, namely $K-y_0$, is a \emph{positive number}. This means the denominator is a strictly decreasing function, so the overall solution is strictly increasing.$

$\noindent If instead $y_0 > K$, then $K-y_0$ is negative, so the denominator is a strictly \emph{increasing} function. This makes the solution a strictly decreasing function.$

1008 · Aug 20, 2016

leehuan said:
I didn't have any problem tackling algebra with this question, but I have something else. (Just going to insert an image here as the question is a bit awkward to type up neatly.)

The answer to (ii) is

$y=\frac{y_0K}{e^{-kt}(K-y_0)-y_0}$

For (iii) and (v), it is quite obvious that y tends to K but I've forgotten how fractions work. Why is it that in (iii) it does so strictly increasing whereas in (iv) it is strictly increasing towards K?

(Cause I'm obviously assuming that there is obviously no need to compute dy/dt=g(t) here)

Wait doesn't y approach -K (not K) as t-> infinity, knowing 0<y_0<K

InteGrand · Aug 20, 2016

1008 said:
Wait doesn't y approach -K (not K) as t-> infinity, knowing 0<y_0<K

It approaches K. leehuan typoed the solution, it should be +y₀ in the denominator. (To see this, note that if we sub. in t = 0, we are supposed to get y = y₀.)

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