How do you solve the differential $\frac{dy}{dx}=\frac{10{x}^{2}}{\sqrt{1+{x}^{3}}}$?

s2vunov
2022-09-29
Answered

How do you solve the differential $\frac{dy}{dx}=\frac{10{x}^{2}}{\sqrt{1+{x}^{3}}}$?

You can still ask an expert for help

recepiamsb

Answered 2022-09-30
Author has **9** answers

We know that

$\frac{d}{dx}\left(\sqrt{1+{x}^{3}}\right)=\frac{3}{2}\frac{{x}^{2}}{\sqrt{1+{x}^{3}}}$

so, grouping variables

$dy=\frac{20}{3}\frac{d}{dx}\left(\sqrt{1+{x}^{3}}\right)dx$ integrating

$y=\frac{20}{3}\sqrt{1+{x}^{3}}+C$

$\frac{d}{dx}\left(\sqrt{1+{x}^{3}}\right)=\frac{3}{2}\frac{{x}^{2}}{\sqrt{1+{x}^{3}}}$

so, grouping variables

$dy=\frac{20}{3}\frac{d}{dx}\left(\sqrt{1+{x}^{3}}\right)dx$ integrating

$y=\frac{20}{3}\sqrt{1+{x}^{3}}+C$

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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.

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What is the solution to the Differential Equation $\frac{4}{{y}^{3}}\frac{dy}{dx}=\frac{1}{x}$?

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We have ${y}^{\prime}=x/y$, which is a first-order homogeneous differential equation.

It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=\pm \sqrt{{x}^{2}+c}$.

Now if we use the substitution $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, and rewrite the differential equation as

${u}^{\prime}x+u=\frac{1}{u}$

and then rearrange to

$\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$

by integrating both parts we get that

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ is undefined. Is this of any significance?

Edit 1:

As pointed out when rearranging from ${u}^{\prime}x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne \pm 1$. Equation (1) does not hold for $u=\pm 1$

Edit 2:

Solving equation (1) for u with $u\ne \pm 1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=\pm 1$

It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=\pm \sqrt{{x}^{2}+c}$.

Now if we use the substitution $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, and rewrite the differential equation as

${u}^{\prime}x+u=\frac{1}{u}$

and then rearrange to

$\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$

by integrating both parts we get that

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ is undefined. Is this of any significance?

Edit 1:

As pointed out when rearranging from ${u}^{\prime}x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne \pm 1$. Equation (1) does not hold for $u=\pm 1$

Edit 2:

Solving equation (1) for u with $u\ne \pm 1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=\pm 1$

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I have this simple differential equation: ${y}^{\prime}=(\mathrm{tan}x)y.$

after integrating $\frac{{y}^{\prime}}{y(x)}=\mathrm{tan}x$

i came up with $\mathrm{log}y(x)=-\mathrm{log}\mathrm{cos}(x)+1.$

now my question is this one: why ${e}^{-\mathrm{log}\mathrm{cos}(x)}=\mathrm{sec}(x)?$

after integrating $\frac{{y}^{\prime}}{y(x)}=\mathrm{tan}x$

i came up with $\mathrm{log}y(x)=-\mathrm{log}\mathrm{cos}(x)+1.$

now my question is this one: why ${e}^{-\mathrm{log}\mathrm{cos}(x)}=\mathrm{sec}(x)?$