How do you find the differential dy of the function $y=3{x}^{\frac{2}{3}}$?

Antwan Perez
2022-10-25
Answered

How do you find the differential dy of the function $y=3{x}^{\frac{2}{3}}$?

You can still ask an expert for help

Plutbantonavv

Answered 2022-10-26
Author has **15** answers

The differential of y is the derivative of the function times the differential of x

$dy=2{x}^{-\frac{1}{3}}dx$

$dy=2{x}^{-\frac{1}{3}}dx$

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I am trying to solve the differential equation

${y}^{\prime}\left(x\right)-\frac{y\left(x\right)}{2x}=x\mathrm{sin}\left(\frac{x}{y\left(x\right)}\right)$

I think it is separable variable differential equations. I tried to substitute:

$z=\frac{x}{y}$

and

${y}^{\prime}=\frac{z-{z}^{\prime}x}{{z}^{2}}$

$\frac{z-{z}^{\prime}x}{{z}^{2}}-\frac{1}{2z}=x\mathrm{sin}(z)$

multiply by ${z}^{2}$

$z-{z}^{\prime}x-\frac{z}{2}={z}^{2}x\mathrm{sin}(z)$

And now I have no idea how to manipulate this.

${y}^{\prime}\left(x\right)-\frac{y\left(x\right)}{2x}=x\mathrm{sin}\left(\frac{x}{y\left(x\right)}\right)$

I think it is separable variable differential equations. I tried to substitute:

$z=\frac{x}{y}$

and

${y}^{\prime}=\frac{z-{z}^{\prime}x}{{z}^{2}}$

$\frac{z-{z}^{\prime}x}{{z}^{2}}-\frac{1}{2z}=x\mathrm{sin}(z)$

multiply by ${z}^{2}$

$z-{z}^{\prime}x-\frac{z}{2}={z}^{2}x\mathrm{sin}(z)$

And now I have no idea how to manipulate this.

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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.

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What is the general solution of the differential equation? : $\frac{dy}{dx}=9{x}^{2}y$

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I have difficulties to solve these two differential equations:

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{{x}^{2}+1}-x$.

2) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. I'm arrived to prove that $\frac{z-1}{(z+1{)}^{3}}={e}^{2c}{x}^{2}$ but I don't know if it's correct. If it's right, how do I explain $z$ to substitute it in $y=xz$? Then, how do I find the constant $c$ ?

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{{x}^{2}+1}-x$.

2) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. I'm arrived to prove that $\frac{z-1}{(z+1{)}^{3}}={e}^{2c}{x}^{2}$ but I don't know if it's correct. If it's right, how do I explain $z$ to substitute it in $y=xz$? Then, how do I find the constant $c$ ?