solve this differential equation

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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Mark Hennings - 7 years, 6 months ago

thanku sir

Sriram Raghavan - 7 years, 4 months ago

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

shiva raj - 7 years, 6 months ago

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

Sriram Raghavan - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$