What is a solution to the differential equation $\frac{dy}{dt}={e}^{t}{(y-1)}^{2}$?

sailorlyts14eh
2022-09-29
Answered

What is a solution to the differential equation $\frac{dy}{dt}={e}^{t}{(y-1)}^{2}$?

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blikath3

Answered 2022-09-30
Author has **7** answers

We have:

$\frac{dy}{dt}={e}^{t}{(y-1)}^{2}$

We can collect terms for similar variables:

$\frac{1}{{(y-1)}^{2}}\frac{dy}{dt}={e}^{t}$

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

$\int \frac{1}{{(y-1)}^{2}}dy=\int {e}^{t}dt$

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

$-\frac{1}{y-1}={e}^{t}+C$

And we can readily rearrange for y:

$-(y-1)=\frac{1}{{e}^{t}+C}$

$\therefore 1-y=\frac{1}{{e}^{t}+C}$

Leading to the General Solution:

$y=1-\frac{1}{{e}^{t}+C}$

$\frac{dy}{dt}={e}^{t}{(y-1)}^{2}$

We can collect terms for similar variables:

$\frac{1}{{(y-1)}^{2}}\frac{dy}{dt}={e}^{t}$

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

$\int \frac{1}{{(y-1)}^{2}}dy=\int {e}^{t}dt$

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

$-\frac{1}{y-1}={e}^{t}+C$

And we can readily rearrange for y:

$-(y-1)=\frac{1}{{e}^{t}+C}$

$\therefore 1-y=\frac{1}{{e}^{t}+C}$

Leading to the General Solution:

$y=1-\frac{1}{{e}^{t}+C}$

asked 2022-02-18

a) Solve the differential equation:

$(x+1)\frac{dy}{dx}-3y={(x+1)}^{4}$

given that$y=16$ and $x=1$ , expressing the answer in the form of $y=f\left(x\right)$ .

b) Hence find the area enclosed by the graphs$y=f\left(x\right),\text{}y={(1-x)}^{4}$ and the $x-a\xi s$ .

I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is:$y={(1+x)}^{4}$ . However how would you calculate the area between the two graphs ($y={(1-x)}^{4}$ and $y={(1+x)}^{4}$ ) and the x-axis.

given that

b) Hence find the area enclosed by the graphs

I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is:

asked 2021-09-11

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.

$f(x,y)={x}^{3}-6xy+8{y}^{3}$

asked 2022-05-15

Im clueless on how to solve the following question...

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

asked 2022-05-21

$g(x)$ is continuous on $[1,2]$ such that $g(1)=0$ and

${g}^{\prime}(x){x}^{2}=\sqrt{1-{g}^{2}(x)}$

Find $g(2)$

I found that $g(x)=\mathrm{sin}(c-\frac{1}{x})$ and since $g(1)=0$ shouldn't my c be equal to 1, so that $\mathrm{sin}(1-1)=0$.

But when I try for $g(2)$ with $c=1$, I am getting a different answer from the textbook.

${g}^{\prime}(x){x}^{2}=\sqrt{1-{g}^{2}(x)}$

Find $g(2)$

I found that $g(x)=\mathrm{sin}(c-\frac{1}{x})$ and since $g(1)=0$ shouldn't my c be equal to 1, so that $\mathrm{sin}(1-1)=0$.

But when I try for $g(2)$ with $c=1$, I am getting a different answer from the textbook.

asked 2022-07-01

How would I solve this differential equation for $y(x)$?

$\frac{dy}{dx}=\frac{y-xy}{x-xy}$

$y-\mathrm{ln}(y)=x-\mathrm{ln}(x)+C$

I'm not sure what to do at this point. I looked it up on WolframAlpha and the solution uses something called a Product Log Function. How does it work? And how does the solution come out to be:

$y(x)=-W(-{e}^{({c}_{1}-x)}x)$

$\frac{dy}{dx}=\frac{y-xy}{x-xy}$

$y-\mathrm{ln}(y)=x-\mathrm{ln}(x)+C$

I'm not sure what to do at this point. I looked it up on WolframAlpha and the solution uses something called a Product Log Function. How does it work? And how does the solution come out to be:

$y(x)=-W(-{e}^{({c}_{1}-x)}x)$

asked 2022-02-15

Most systems can be modeled using first order system just like a temperature control system of an incubator.

Question 1: How can we know that a certain system can be modeled using first order differential equation?

${a}_{1}\frac{dy}{dt}+{a}_{0}y=F\left(t\right)\text{}\text{}\text{}\text{}\left(1\right)$

A first order differential equation in Equation (1) may be written as:

$\tau \frac{dy}{dt}+y=kF\left(t\right)\text{}\text{}\text{}\text{}\left(2\right)$

Question 2: If a step input

$F\left(t\right)=\{\begin{array}{cc}0& \text{}\text{}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t0\\ A& \text{}\text{}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t0\end{array}\text{}\text{}\text{}\text{}\left(3\right)$

of the system has an output response of a

$y\left(t\right)=kA(1-{e}^{\frac{-t}{\tau}}\text{}\text{}\text{}\text{}\left(4\right)$

then the system can be modeled as first differential equation where

${y}_{max}=kA\text{}\text{}\text{}\text{}\left(5\right)$

Is this correct?

Given a step input and the graph of the step response of the system. Am i correct if I say that if the output of the system behaves similar to equation (4), then I can model the system using first differential equation?

Question 1: How can we know that a certain system can be modeled using first order differential equation?

A first order differential equation in Equation (1) may be written as:

Question 2: If a step input

of the system has an output response of a

then the system can be modeled as first differential equation where

Is this correct?

Given a step input and the graph of the step response of the system. Am i correct if I say that if the output of the system behaves similar to equation (4), then I can model the system using first differential equation?

asked 2021-01-28

Solve differential equation ${y}^{\prime}+2xy=4x$