Solve differential equation

Chaya Galloway
2021-03-07
Answered

Solve differential equation

You can still ask an expert for help

mhalmantus

Answered 2021-03-08
Author has **106** answers

Integrating on both sides

Differentiating

Jeffrey Jordon

Answered 2021-12-25
Author has **2581** answers

Answer is given below (on video)

asked 2022-09-04

What is the general solution of the differential equation? : $y\prime =x(1+{y}^{2})$

asked 2022-07-10

I have a first order linear differential equation (a variation on a draining mixing tank problem) with many constants, and want to separate variables to solve it.

$\frac{dy}{dt}={k}_{1}+{k}_{2}\frac{y}{{k}_{3}+{k}_{4}t}$

y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, ${k}_{1}$ through ${k}_{4}$.

Separation of variables is made difficult by ${k}_{1}$, and I've considered an integrating factor, but think I might be missing something simple.

$\frac{dy}{dt}={k}_{1}+{k}_{2}\frac{y}{{k}_{3}+{k}_{4}t}$

y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, ${k}_{1}$ through ${k}_{4}$.

Separation of variables is made difficult by ${k}_{1}$, and I've considered an integrating factor, but think I might be missing something simple.

asked 2022-05-17

We have ${y}^{\prime}=x/y$, which is a first-order homogeneous differential equation.

It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=\pm \sqrt{{x}^{2}+c}$.

Now if we use the substitution $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, and rewrite the differential equation as

${u}^{\prime}x+u=\frac{1}{u}$

and then rearrange to

$\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$

by integrating both parts we get that

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ is undefined. Is this of any significance?

Edit 1:

As pointed out when rearranging from ${u}^{\prime}x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne \pm 1$. Equation (1) does not hold for $u=\pm 1$

Edit 2:

Solving equation (1) for u with $u\ne \pm 1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=\pm 1$

It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=\pm \sqrt{{x}^{2}+c}$.

Now if we use the substitution $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, and rewrite the differential equation as

${u}^{\prime}x+u=\frac{1}{u}$

and then rearrange to

$\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$

by integrating both parts we get that

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ is undefined. Is this of any significance?

Edit 1:

As pointed out when rearranging from ${u}^{\prime}x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne \pm 1$. Equation (1) does not hold for $u=\pm 1$

Edit 2:

Solving equation (1) for u with $u\ne \pm 1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=\pm 1$

asked 2022-09-11

How to you find the general solution of $\frac{dy}{dx}={\mathrm{tan}}^{2}x$?

asked 2021-01-19

A tank contains 70 gallons of pure water. A brine solution with 2 lbs/gal of salt enters at 2.25 gal/min and the well - stirred mixture leaves at the same rate. Find the amount of salt in the tank at any time and the time at which the brine leaving the tank will contain 15 lbs of salt.

asked 2022-11-25

How can I solve this differential equation? : $xydx-({x}^{2}+1)dy=0$

asked 2020-11-14

Solve and classify equation
$dy/dx=-({e}^{x}+y)/(k+x+y{e}^{x})$ , y(0)=1