What is a solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{sin}x}{y}$?

tophergopher3wo
2022-09-13
Answered

What is a solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{sin}x}{y}$?

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asked 2022-07-03

Given the following differential equation i have to find the general solution using the integrating factor technique.

$\frac{dy}{dx}+\frac{y}{x}={e}^{x}$

I know that P(x)=$\frac{1}{x}$ and Q(x)=${e}^{x}$ also i found $\mu (x)={e}^{lnx}$

Using the formula

$y=\frac{1}{\mu (x)}\int \mu (x)Q(x)dx$

i found this general solution:

$\frac{{e}^{x}x}{x}-\frac{{e}^{x}}{x}+\frac{C}{x}$

Help, maybe this is incorrect

$\frac{dy}{dx}+\frac{y}{x}={e}^{x}$

I know that P(x)=$\frac{1}{x}$ and Q(x)=${e}^{x}$ also i found $\mu (x)={e}^{lnx}$

Using the formula

$y=\frac{1}{\mu (x)}\int \mu (x)Q(x)dx$

i found this general solution:

$\frac{{e}^{x}x}{x}-\frac{{e}^{x}}{x}+\frac{C}{x}$

Help, maybe this is incorrect

asked 2022-06-21

I have a differential equation of the form

$dy/dx+p(x)y=q(x)$

under the condition that $q(x)=300$ if $y<3312$ and $q(x)=0$ if $y\ge 3312$.

I could not understand how to solve this differential equation with such heavy side function ?

Any hints?

$dy/dx+p(x)y=q(x)$

under the condition that $q(x)=300$ if $y<3312$ and $q(x)=0$ if $y\ge 3312$.

I could not understand how to solve this differential equation with such heavy side function ?

Any hints?

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.

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