Consider the equation $dy/dx={y}^{1/3}$.

Dean Summers
2022-07-25
Answered

Consider the equation $dy/dx={y}^{1/3}$.

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nuramaaji2000fh

Answered 2022-07-26
Author has **18** answers

Given that $\frac{dy}{dx}={y}^{\frac{1}{3}}$

$\Rightarrow \frac{dy}{{y}^{\frac{1}{3}}}dx$

$\Rightarrow {y}^{-\frac{1}{3}}dy=dx$

$\Rightarrow \int {y}^{-\frac{1}{3}}dy=\int dx$

$\Rightarrow \frac{{y}^{1-\frac{1}{3}}}{1-\frac{1}{3}}=x+C$

$\Rightarrow {y}^{\frac{2}{3}}=\frac{2}{3}x+C$

$\Rightarrow y=(\frac{2}{3}x+C{)}^{\frac{3}{2}}$

$\Rightarrow \frac{dy}{{y}^{\frac{1}{3}}}dx$

$\Rightarrow {y}^{-\frac{1}{3}}dy=dx$

$\Rightarrow \int {y}^{-\frac{1}{3}}dy=\int dx$

$\Rightarrow \frac{{y}^{1-\frac{1}{3}}}{1-\frac{1}{3}}=x+C$

$\Rightarrow {y}^{\frac{2}{3}}=\frac{2}{3}x+C$

$\Rightarrow y=(\frac{2}{3}x+C{)}^{\frac{3}{2}}$

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I would like to solve:

$y-{y}^{\prime}x-{y}^{\prime 2}=0$

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

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That is, we have: $x=-2dy/dx$. Thus, we obtain

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Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

$y-{y}^{\prime}x-{y}^{\prime 2}=0$

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

$x=-2t$

That is, we have: $x=-2dy/dx$. Thus, we obtain

$y=-\frac{{x}^{2}}{4}+C$

Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

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with$y\left(0\right)=a\text{}\text{and}\text{}{y}^{\prime}\left(0\right)=0$

Where a is a known constant.

with

Where a is a known constant.

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