Question

A Bernoulli differential equation is one of the form

Answer 1

$y'-\dfrac5xy=\dfrac{y^5}{x^9}$

Divide through both sides by $y^5$:

$y^{-5}y'-\dfrac5xy^{-4}=x^{-9}$

Now let $z=y^{-4}$, so that $z'=-4y^{-5}y'$. Then

$-\dfrac{z'}4-\dfrac5xz=x^{-9}$

$z'+\dfrac{20}xz=-4x^{-9}$

Multiply both sides by $x^{20}$:

$x^{20}z'+20x^{19}z=\left(x^{20}z\right)'=-4x^{11}$

Integrate both sides to get

$x^{20}z=-\dfrac{x^{12}}3+C\implies z=-\dfrac1{3x^8}+\dfrac C{x^{20}}$

Solve for $y$:

$y=\left(\dfrac C{x^{20}}-\dfrac1{3x^8}\right)^{-1/4}$

Given that $y(1)=1$, we have

$1=\left(C-\dfrac13\right)^{-1/4}\implies C=\dfrac43$

so the particular solution is

$y=\left(\dfrac4{3x^{20}}-\dfrac1{3x^8}\right)^{-1/4}$

which we can rewrite as

$\boxed{y=\dfrac{3^{1/4}}{x^5\left(4-x^{12}\right)^{1/4}}}$