What is a solution to the differential equation $\frac{dy}{dx}=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}$?

rustenig
2022-09-14
Answered

What is a solution to the differential equation $\frac{dy}{dx}=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}$?

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Ashlee Ramos

Answered 2022-09-15
Author has **20** answers

The equation simplifies:

$\frac{dy}{dx}=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)$

Separate variables:

$dy=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$

Integrate:

$\int dy=2\int \mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$

$y=2\mathrm{ln}\left|\mathrm{sin}\left(x\right)\right|-2\mathrm{ln}\left|\mathrm{cos}\left(x\right)\right|+C$

$\frac{dy}{dx}=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)$

Separate variables:

$dy=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$

Integrate:

$\int dy=2\int \mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$

$y=2\mathrm{ln}\left|\mathrm{sin}\left(x\right)\right|-2\mathrm{ln}\left|\mathrm{cos}\left(x\right)\right|+C$

equipokypip1

Answered 2022-09-16
Author has **5** answers

More details

First we should separate the variables, which means that we can treat $\frac{dy}{dx}$ like division. We can move the dx to the right hand side of the equation to be with all the other terms including x.

$dy=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

Now integrate both sides:

$\int dy=2\int \frac{{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

On the right hand side, let u=cotx. This implies that $du=-{\mathrm{csc}}^{2}xdx$.

$y=-2\int \frac{-{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

$y=-2\int \frac{du}{u}$

$y=-2\mathrm{ln}\left|u\right|+C$

$y=-2\mathrm{ln}\left|\mathrm{cot}x\right|+C$

One possible simplification we could make if we wanted would be to bring the −1 outside the logarithm into the logarithm as a −1 power:

$y=2\mathrm{ln}\left|\mathrm{tan}x\right|+C$

First we should separate the variables, which means that we can treat $\frac{dy}{dx}$ like division. We can move the dx to the right hand side of the equation to be with all the other terms including x.

$dy=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

Now integrate both sides:

$\int dy=2\int \frac{{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

On the right hand side, let u=cotx. This implies that $du=-{\mathrm{csc}}^{2}xdx$.

$y=-2\int \frac{-{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$

$y=-2\int \frac{du}{u}$

$y=-2\mathrm{ln}\left|u\right|+C$

$y=-2\mathrm{ln}\left|\mathrm{cot}x\right|+C$

One possible simplification we could make if we wanted would be to bring the −1 outside the logarithm into the logarithm as a −1 power:

$y=2\mathrm{ln}\left|\mathrm{tan}x\right|+C$

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I have the following differential equation

$s(1-s)t={f}^{\prime}(t)(f(t)-st)$

Initial condition: $f(0)=0.$

I solved it and got the solution as

$f(t)=\sqrt{2{c}_{1}+{s}^{2}-2(s-1)s\mathrm{ln}(t)}+s$

But the answer given is

$f(t)=k(s)t,$

where

$k(s)=\frac{s+\sqrt{4s-3{s}^{2}}}{2}.$

If anyone can provide me some hint on how to proceed and reach the specified answer, I would be really grateful.

$s(1-s)t={f}^{\prime}(t)(f(t)-st)$

Initial condition: $f(0)=0.$

I solved it and got the solution as

$f(t)=\sqrt{2{c}_{1}+{s}^{2}-2(s-1)s\mathrm{ln}(t)}+s$

But the answer given is

$f(t)=k(s)t,$

where

$k(s)=\frac{s+\sqrt{4s-3{s}^{2}}}{2}.$

If anyone can provide me some hint on how to proceed and reach the specified answer, I would be really grateful.

asked 2022-06-23

I am studying numeric solutions of differential equations, and part of my reading is found in Simmonds' book, Differential Equations with Applications and Historical Notes. Although the chapter on numerical methods is written by John S. Robertson.

In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?

For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value $y(a)=\alpha $

These are then transformed into the integral equation

$y({x}_{1})-y({x}_{0})={\int}_{{x}_{0}}^{{x}_{1}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\text{.}$

Then, using Taylor's theorem and substituting back into that formula, the error is found to be

$\frac{{h}^{2}}{2}{y}^{\u2033}(\xi )\text{,}$

which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is ${y}^{\u2033}(\xi )$. I don't understand why this is necessarily true.

In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?

For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value $y(a)=\alpha $

These are then transformed into the integral equation

$y({x}_{1})-y({x}_{0})={\int}_{{x}_{0}}^{{x}_{1}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\text{.}$

Then, using Taylor's theorem and substituting back into that formula, the error is found to be

$\frac{{h}^{2}}{2}{y}^{\u2033}(\xi )\text{,}$

which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is ${y}^{\u2033}(\xi )$. I don't understand why this is necessarily true.