Chardonnay Felix
2021-02-06
Answered

Solve differential equation${y}^{\prime}(t)=-3y+9$ , y(0)=4

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Brittany Patton

Answered 2021-02-07
Author has **100** answers

now as y(0)=4

C= 4-27

C= -23

C= -23

Jeffrey Jordon

Answered 2021-12-25
Author has **2262** answers

Answer is given below (on video)

asked 2020-11-14

Solve differential equation

asked 2022-01-20

Solve the first-order differential equation
State which method you are using!

${x}^{2}{y}^{2}dx-({x}^{3}+1)dy=0$

asked 2021-12-30

Brine containing 3lb/gal of salt enters a tank at the rate of 4gal/min and the solution well-stirred leaves at 2gal/min. The tank initially contains 20 gal of water. What is the amount of salt in the tank after 5 minutes?

asked 2021-12-18

Topic:

Exact Differential Equations

Instructions:

Solve the Exact Differential Equations of GIven Problem #2 only, ignore the other given problem.

Note:

Please answer all of the problems in the attached photo. I will rate you with “like/upvote” after; if you answer all of the problems. If not, I will rate you “unlike/downvote.” Kindly show the complete step-by-step solution.

1.$(x{e}^{x}y+{e}^{x}y)dx+(8+x{e}^{x})dy=0;y\left(0\right)=-4$

2.$\left(\frac{x+y}{{y}^{2}+1}\right)dy+(\mathrm{arctan}y+x)dx=0$

Exact Differential Equations

Instructions:

Solve the Exact Differential Equations of GIven Problem #2 only, ignore the other given problem.

Note:

Please answer all of the problems in the attached photo. I will rate you with “like/upvote” after; if you answer all of the problems. If not, I will rate you “unlike/downvote.” Kindly show the complete step-by-step solution.

1.

2.

asked 2021-09-13

Obtain Laplace transforms for the function $t}^{2$ step(t-2)

asked 2022-01-15

A question asks us to solve the differential equation

$-u\left(x\right)=\delta \left(x\right)$

with boundary conditions

$u(-2)=0\text{}\text{and}\text{}u\left(3\right)=0$ where $\delta \left(x\right)$ is the Dirac delta function. But inside the same question, teacher gives the solution in two pieces as $u=A(x+2)\text{}\text{for}\text{}x\le 0\text{}\text{and}\text{}u=B(x-3)\text{}\text{for}\text{}x\ge 0$ . I understand when we integrate the delta function twice the result is the ramp function R(x). However elsewhere in his lecture the teacher had given the general solution of that DE as

$u\left(x\right)=-R\left(x\right)+C+Dx$

with boundary conditions

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 $u=\pm 1$