What is the general solution of the differential equation $\frac{dy}{dx}+y=x{y}^{3}$?

redolrn
2022-09-27
Answered

What is the general solution of the differential equation $\frac{dy}{dx}+y=x{y}^{3}$?

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Derick Ortiz

Answered 2022-09-28
Author has **11** answers

Making the change of variable $y=\frac{1}{z}$ we have the new version

$\frac{dy}{dx}+y-x{y}^{3}=0\to \frac{x-{z}^{2}+zz\prime}{{z}^{3}}=0$ or

$x-{z}^{2}+zz\prime =0$

Now calling $\xi ={z}^{2}$ we have

$x-\xi +\frac{1}{2}\xi \prime =0$

Solving for $\xi$ we obtain easily

$\xi =\frac{1}{2}+x+C{e}^{2x}={z}^{2}$ then

$z=\pm \sqrt{\frac{1}{2}+x+C{e}^{2x}}=\frac{1}{y}$ then finally

$y=\pm \frac{1}{\sqrt{\frac{1}{2}+x+C{e}^{2x}}}$

$\frac{dy}{dx}+y-x{y}^{3}=0\to \frac{x-{z}^{2}+zz\prime}{{z}^{3}}=0$ or

$x-{z}^{2}+zz\prime =0$

Now calling $\xi ={z}^{2}$ we have

$x-\xi +\frac{1}{2}\xi \prime =0$

Solving for $\xi$ we obtain easily

$\xi =\frac{1}{2}+x+C{e}^{2x}={z}^{2}$ then

$z=\pm \sqrt{\frac{1}{2}+x+C{e}^{2x}}=\frac{1}{y}$ then finally

$y=\pm \frac{1}{\sqrt{\frac{1}{2}+x+C{e}^{2x}}}$

asked 2022-09-27

Solve the differential equation $\frac{dy}{dx}+y=0$ ?

asked 2022-06-16

I am familiar with first-order differential equations and how to solve them, but only for the case when there is a single unknown function.

For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:

$\begin{array}{}\text{(1)}& {f}^{\prime}(t)=-Af(t)+Bg(t)\end{array}$

$\begin{array}{}\text{(2)}& {g}^{\prime}(t)=-Bg(t)+Af(t)\end{array}$

Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: $f(0)=0$ and $g(0)=1$.

I already have knowledge of the answer for f(t), which is:

$\begin{array}{}\text{(3)}& f(t)=\frac{B}{A+B}(1-{e}^{-(A+B)t})\end{array}$

I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.

Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.

For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:

$\begin{array}{}\text{(1)}& {f}^{\prime}(t)=-Af(t)+Bg(t)\end{array}$

$\begin{array}{}\text{(2)}& {g}^{\prime}(t)=-Bg(t)+Af(t)\end{array}$

Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: $f(0)=0$ and $g(0)=1$.

I already have knowledge of the answer for f(t), which is:

$\begin{array}{}\text{(3)}& f(t)=\frac{B}{A+B}(1-{e}^{-(A+B)t})\end{array}$

I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.

Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.

asked 2021-03-02

Transform the given initial value problem into an initial value problem for two first-order quations

asked 2022-06-22

I have a linear first order ordinary differential equation

$\frac{dy}{dx}+\mathrm{tan}(x)y=2{\mathrm{cos}}^{2}x\mathrm{sin}x-\mathrm{sec}x$

with an initial condition as $y(\frac{\pi}{4})=3\sqrt{2}$

My integrating factor $\mu (x)=\mathrm{sec}x$

After multiplication with the integration factor what I get is:

$(secx\text{}y{)}^{\prime}=sin2x-se{c}^{2}x$

or

$(secx\text{}y{)}^{\prime}=2sinxcosx-se{c}^{2}x$

If I use the first equation I get:

$y(x)=\frac{\frac{1}{2}cos2x-tanx+c}{secx}$

and using the second equation I get:

$y(x)=\frac{si{n}^{2}x-tanx+c}{secx}$

$\int 2\text{}sinx\text{}cosx\text{}dx=2\frac{si{n}^{2}x}{2}=si{n}^{2}x$

with the first equation I get c=7 and second equation I get $c=\frac{13}{2}$.

It is a very simple differential equation but when I solve it I get two different answers. Is this ok?

$\frac{dy}{dx}+\mathrm{tan}(x)y=2{\mathrm{cos}}^{2}x\mathrm{sin}x-\mathrm{sec}x$

with an initial condition as $y(\frac{\pi}{4})=3\sqrt{2}$

My integrating factor $\mu (x)=\mathrm{sec}x$

After multiplication with the integration factor what I get is:

$(secx\text{}y{)}^{\prime}=sin2x-se{c}^{2}x$

or

$(secx\text{}y{)}^{\prime}=2sinxcosx-se{c}^{2}x$

If I use the first equation I get:

$y(x)=\frac{\frac{1}{2}cos2x-tanx+c}{secx}$

and using the second equation I get:

$y(x)=\frac{si{n}^{2}x-tanx+c}{secx}$

$\int 2\text{}sinx\text{}cosx\text{}dx=2\frac{si{n}^{2}x}{2}=si{n}^{2}x$

with the first equation I get c=7 and second equation I get $c=\frac{13}{2}$.

It is a very simple differential equation but when I solve it I get two different answers. Is this ok?

asked 2022-09-23

What is a particular solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{ln}x}{xy}$ and y(1)=2?

asked 2022-07-14

A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) starts to decrease according to the formula

$\frac{dv}{dt}=-kv\phantom{\rule{1em}{0ex}}(1)$

where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution

$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=\frac{dv}{ds}v$

to solve the differential equation.

Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.

What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).

So far I've found velocity as a function of displacement (v(s)):

$\frac{dv}{dt}=-k\frac{ds}{dt}\phantom{\rule{1em}{0ex}}\text{(from (1))}$

$\int \frac{dv}{dt}dt=-k\int \frac{ds}{dt}dt$

$v(s)=-ks+C$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=100,\text{}v(1)=50\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k=50$

$v(s)=-50s+100$

Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).

The answer key says $v(t)=100{e}^{-50t}$ and $s(t)=2(1-{e}^{-50t})$

I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.

$\frac{dv}{dt}=-kv\phantom{\rule{1em}{0ex}}(1)$

where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution

$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=\frac{dv}{ds}v$

to solve the differential equation.

Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.

What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).

So far I've found velocity as a function of displacement (v(s)):

$\frac{dv}{dt}=-k\frac{ds}{dt}\phantom{\rule{1em}{0ex}}\text{(from (1))}$

$\int \frac{dv}{dt}dt=-k\int \frac{ds}{dt}dt$

$v(s)=-ks+C$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=100,\text{}v(1)=50\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k=50$

$v(s)=-50s+100$

Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).

The answer key says $v(t)=100{e}^{-50t}$ and $s(t)=2(1-{e}^{-50t})$

I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.

asked 2022-09-28

What is a solution to the differential equation $\frac{dx}{dt}=t(x-2)$?