-10.5.99 separable eq y'=x^4y^4

[karush]

Solve the separable differential equation
$\displaystyle y'=x^4y^4$
Solve for $y$ if possible.

$\displaystyle
y=\frac{{y'}^{(1/4)}}{x}$
Not sure ?

 

[Greg]

$\frac{dy}{dx}=x^4y^4$

$\frac{1}{y^4}\,dy=x^4\,dx$

$\int\frac{1}{y^4}\,dy=\int x^4\,dx$

 

[Greg]

$$\frac{dy}{dx}=x^4y^4$$

$$\frac{1}{y^4}\,dy=x^4\,dx$$

$$\int\frac{1}{y^4}\,dy=\int x^4\,dx$$

$$-\frac13y^{-3}=\frac15x^5+c_1$$

$$y^{-3}=-\frac35\left(x^5+c_2\right)$$

$$y=\left[-\frac35\left(x^5+c_2\right)\right]^{-1/3}$$

$$y=-\left[\frac35\left(x^5+c_2\right)\right]^{-1/3}$$

 

[karush]

Where does the $-\frac{1}{3}$ inside the $\left[\right]$ come from

 

[Greg]

It comes from a typo. It should be $-\frac35$.

 

[MarkFL]

During the process of separation of variables (dividing through by $y^4$), the trivial solution:

\(\displaystyle y\equiv0\)

was lost. :)