## anonymous 5 years ago Solve this DE: x(dy/dx)=1/(y^3)

1. anonymous

$y^3dy=\frac1xdx$

2. anonymous

$x\frac{dy}{dx} = \frac{1}{y^3}$ multiply by y^3, multiply by dx and divide by x to get: $y^3dy = \frac{1}{x}dx$ Now, as if by magic, integrate! $\displaystyle\int y^3dy = \displaystyle\int x^{-1}dx$ and remember your +C

3. anonymous

$\int\limits y^3dy=\int\limits \frac 1xdx$ $\frac {y^4}4=\ln x+c$ $y=(\ln x^4+c)^{1/4}$

4. anonymous

OOh thanks for the +C reminder or else i was trying to cancel out the exponent.. X_X

5. anonymous

$y(x) = \left(4\ln x + 4c\right)^{\frac{1}{4}}$ and let $4c = C \in \mathbb{R}$