What is the general solution of the differential equation $\frac{dy}{dx}=8\mathrm{sin}2x$?

Milton Anderson
2022-09-12
Answered

What is the general solution of the differential equation $\frac{dy}{dx}=8\mathrm{sin}2x$?

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Solve the differential equation $2x\mathrm{ln}x\frac{dy}{dx}+y=0$?

asked 2021-09-06

Find all the second partial derivatives

$v=\frac{xy}{x-y}$

asked 2022-02-15

Given that $y=\mathrm{sin}\left(x\right)$ is an expicit function of the first-order differential equation $\frac{dy}{dx}=\sqrt{1-{y}^{2}}$ . Find an interval I of definition, the solution interval.

So I got to the point that$\mathrm{cos}\left(x\right)\frac{=}{\mathrm{cos}\left(x\right)}$ , what do I do next?

So I got to the point that

asked 2022-06-03

y' + (3/x ) = x-1 , y = x^2 /5 - x/4 + 2/x^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 2022-05-13

In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

asked 2022-01-20

Solve the first order differential equation using any acceptable method.

$\mathrm{sin}\left(x\right)\frac{dy}{dx}+\left(\mathrm{cos}\left(x\right)\right)y=0$ , $y\left(\frac{7\pi}{6}\right)=-2$