# Math Help - solving differential equation using substution

1. ## solving differential equation using substution

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!

2. Originally Posted by Niall101
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 $z = 7x - 3y$, note that

$\frac{dz}{dx} = \frac{d}{dx}(7x - 3y)$

$\frac{dz}{dx} = 7 - 3\frac{dy}{dx}$.

So $\frac{dy}{dx} = \frac{1}{3}\left(7 - \frac{dz}{dx}\right)$.

$\frac{dy}{dx} = (7x - 3y)^3$

$\frac{1}{3}\left(7 - \frac{dz}{dx}\right) = z^3$

$7 - \frac{dz}{dx} = 3z^3$

$\frac{dz}{dx} = 7 - 3z^3$

$\frac{dx}{dz} = \frac{1}{7 - 3z^3}$.

You can now solve this using partial fractions.

3. The roots of the polynomial $z^{3}- 7$ are...

$z_{0} = \sqrt[3]{7}$

$z_{1} = \sqrt[3]{7}\cdot e^{i \frac{2}{3} \pi}$

$z_{2} = \sqrt[3]{7}\cdot e^{i \frac{4}{3} \pi}$ (1)

... so that is...

$\frac{1}{z^{3}-7} = \frac{a_{0}}{z-z_{0}} + \frac{a_{1}}{z-z_{1}} + \frac{a_{2}}{z-z_{2}}$ (2)

... where...

$a_{k} = \lim_{z \rightarrow z_{k}} \frac{z-z_{k}}{z^{3} - 7}, k=0,1,2$ (3)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by Prove It
If you are using the substitution $z = 7x - 3y$, note that

$\frac{dz}{dx} = \frac{d}{dx}(7x - 3y)$

$\frac{dz}{dx} = 7 - 3\frac{dy}{dx}$.

So $\frac{dy}{dx} = \frac{1}{3}\left(7 - \frac{dz}{dx}\right)$.

$\frac{dy}{dx} = (7x - 3y)^3$

$\frac{1}{3}\left(7 - \frac{dz}{dx}\right) = z^3$

$7 - \frac{dz}{dx} = 3z^3$

$\frac{dz}{dx} = 7 - 3z^3$

$\frac{dx}{dz} = \frac{1}{7 - 3z^3}$.

You can now solve this using partial fractions.
Hey thank you! I had this on one of my attempts with disregarded it as i ended up with a horrible looking solution. Thatnks so much!