New user? Sign up

Existing user? Log in

y(x+y^3)dx=x(y^3-x)dy.

Note by Sriram Raghavan 4 years, 6 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

For the differential equation \[ \frac{dy}{dx} \; = \; \frac{y(x+y^3)}{x(y^3-x)} \] try the substitution \(y \,=\, x^{\frac13}u\), so that \[ \begin{array}{rcl}\displaystyle x^{\frac13}\frac{du}{dx} + \tfrac13x^{-\frac23}u & = & \displaystyle\frac{x^{\frac13}u(x + xu^3)}{x(xu^3-x)} \; = \; x^{-\frac23}\frac{u(u^3+1)}{u^3-1} \\ xu' + \tfrac13u & = & \displaystyle\frac{u(u^3+1)}{u^3-1} \\ xu' & = & \displaystyle\frac{2u(u^3+2)}{3(u^3-1)} \\ \displaystyle\int \Big(\frac{3u^2}{u^3+2} - \frac{1}{u}\Big)\,du \; = \; \int \frac{2(u^3-1)}{u(u^3+2)}\,du & = & \displaystyle\int \frac{4}{3x}\,dx \; = \; \tfrac43\ln x + c \\ \displaystyle \ln \Big(\frac{u^3+2}{u}\Big) & = & \tfrac43\ln x + c \\ [2ex] \displaystyle \frac{u^3+2}{u} & = & Ax^{\frac43} \\ u^3 + 2 & = & Aux^{\frac43} \\ xu^3 + 2x & = & Aux^{\frac73} \\ y^3 + 2x & = & Ax^2y \end{array} \]

Log in to reply

thanku sir

I think the answer is x=(-y^3)

i think tis is nt the answer ,,,the answer is given in the book..bt no steps..@shiva raj

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestFor the differential equation \[ \frac{dy}{dx} \; = \; \frac{y(x+y^3)}{x(y^3-x)} \] try the substitution \(y \,=\, x^{\frac13}u\), so that \[ \begin{array}{rcl}\displaystyle x^{\frac13}\frac{du}{dx} + \tfrac13x^{-\frac23}u & = & \displaystyle\frac{x^{\frac13}u(x + xu^3)}{x(xu^3-x)} \; = \; x^{-\frac23}\frac{u(u^3+1)}{u^3-1} \\ xu' + \tfrac13u & = & \displaystyle\frac{u(u^3+1)}{u^3-1} \\ xu' & = & \displaystyle\frac{2u(u^3+2)}{3(u^3-1)} \\ \displaystyle\int \Big(\frac{3u^2}{u^3+2} - \frac{1}{u}\Big)\,du \; = \; \int \frac{2(u^3-1)}{u(u^3+2)}\,du & = & \displaystyle\int \frac{4}{3x}\,dx \; = \; \tfrac43\ln x + c \\ \displaystyle \ln \Big(\frac{u^3+2}{u}\Big) & = & \tfrac43\ln x + c \\ [2ex] \displaystyle \frac{u^3+2}{u} & = & Ax^{\frac43} \\ u^3 + 2 & = & Aux^{\frac43} \\ xu^3 + 2x & = & Aux^{\frac73} \\ y^3 + 2x & = & Ax^2y \end{array} \]

Log in to reply

thanku sir

Log in to reply

I think the answer is x=(-y^3)

Log in to reply

i think tis is nt the answer ,,,the answer is given in the book..bt no steps..@shiva raj

Log in to reply