Solving the differential equation $(xy^{4}+y)dx - xdy = 0$

Question

The differential equation is :

$(xy^{4}+y)dx - xdy = 0$

I am trying to simplify the equation to the form $\dfrac{fdg-gdf}{f^{2}}$ so that I can reduce it to $d(\dfrac{g}{f})$ but I am unable to do it.

Any ideas are appreciated.

Answer 1

You can do it ...... $$(xy^{4}+y)dx - xdy = 0$$ $$(xy^{4}+y)= xy' $$ $$xy^{4}= -(y-xy') $$ Here you have what you wanted $$xy^{2}= -\left(\frac {y-xy' }{y^2}\right )$$ $$xy^{2}= -\left(\frac {x }{y}\right)' \implies \int x^{3}dx= -\int\left(\frac xy \right )^2d\left(\frac {x }{y}\right)$$ You can conitnue but I prefer to substitute $\frac xy=z$ $$xy^{2}= -\left(\frac {x }{y}\right)' \implies \frac {x^{3}} {z^2}= -z'$$ Which is easy to integrate $$\int x^{3}dx= -\int z^2 dz \implies \frac {x^4}4+K=- \frac {z^3}3 \implies y^3(\frac {x^4}4+K)=- \frac {x^3}3 $$ $$\boxed {y^3(x)=- \frac {4x^3}{({3x^4}+K)}} $$

Answer 2

If you instead let $y(x)=x u(x)$ your differential equation takes the form $$ \bigl(x(xu)^4+xu)\bigr)\,dx-x(x\,du+u\,dx)=0 $$ that is $$ x^5u^4\,dx=x^2\,du. $$ I'm sure you can solve this differential equation.

Answer 3

Rewrite the equation like this $$ \frac{dy}{dx} - \frac{1}{x}y = y^4 $$

This is a Bernoulli equation. Divide through by $y^4$

$$ y^{-4}\frac{dy}{dx} - \frac{1}{x}y^{-3} = 1$$

Make the substitution $z = y^{-3}$ to obtain

$$ \frac{dz}{dx} + \frac{3}{x}z = -3 $$

This is now just a regular linear first-order equation. The integrating factor is $\mu = e^{3\ln x} = x^3$, you can do the rest