DY/DX = (7X - 3Y)^3
I used substitution Z = (7X -3Y) Then seperate the variables but i cannot do the integration of 1/(Z^3 - 7) Unless i am supposed to use another substitution??
Thanks!
If you are using the substitution $\displaystyle z = 7x - 3y$, note that
$\displaystyle \frac{dz}{dx} = \frac{d}{dx}(7x - 3y)$
$\displaystyle \frac{dz}{dx} = 7 - 3\frac{dy}{dx}$.
So $\displaystyle \frac{dy}{dx} = \frac{1}{3}\left(7 - \frac{dz}{dx}\right)$.
So from your original DE
$\displaystyle \frac{dy}{dx} = (7x - 3y)^3$
$\displaystyle \frac{1}{3}\left(7 - \frac{dz}{dx}\right) = z^3$
$\displaystyle 7 - \frac{dz}{dx} = 3z^3$
$\displaystyle \frac{dz}{dx} = 7 - 3z^3$
$\displaystyle \frac{dx}{dz} = \frac{1}{7 - 3z^3}$.
You can now solve this using partial fractions.
The roots of the polynomial $\displaystyle z^{3}- 7$ are...
$\displaystyle z_{0} = \sqrt[3]{7}$
$\displaystyle z_{1} = \sqrt[3]{7}\cdot e^{i \frac{2}{3} \pi}$
$\displaystyle z_{2} = \sqrt[3]{7}\cdot e^{i \frac{4}{3} \pi}$ (1)
... so that is...
$\displaystyle \frac{1}{z^{3}-7} = \frac{a_{0}}{z-z_{0}} + \frac{a_{1}}{z-z_{1}} + \frac{a_{2}}{z-z_{2}} $ (2)
... where...
$\displaystyle a_{k} = \lim_{z \rightarrow z_{k}} \frac{z-z_{k}}{z^{3} - 7}, k=0,1,2$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$