Question

Use an appropriate substitution to solve the equation

Answer 1

$y'-\dfrac8xy=\dfrac{y^5}{x^{20}}$

This ODE is in standard Bernoulli form, so we can divide through both sides by $y^5$ to get

$y^{-5}y'-\dfrac8xy^{-4}=\dfrac1{x^{20}}$

Substitute $z=y^{-4}$, so that $z'=-4y^{-5}y'$. Then the ODE can be written as a linear one in $z$:

$-\dfrac14z'-\dfrac8xz=\dfrac1{x^{20}}$
$z'+\dfrac{32}xz=-\dfrac4{x^{20}}$

Multiplying both sides by $x^{32}$ gives

$x^{32}z'+32x^{31}z=-4x^{12}$
$(x^{32}z)'=-4x^{12}$
$x^{32}z=\displaystyle-4\int x^{12}\,\mathrm dx$
$x^{32}z=-\dfrac4{13}x^{13}+C$
$z=-\dfrac4{13x^{19}}+\dfrac C{x^{32}}$

Back-substitute to solve for $y$:

$\dfrac1{y^4}=-\dfrac4{13x^{19}}+\dfrac C{x^{32}}$
$y^4=\dfrac1{Cx^{-32}-\frac4{13}x^{-19}}$
$y=\left(\dfrac{x^{32}}{C-\frac4{13}x^{13}}\right)^{1/4}$
$y=\dfrac{x^8}{(C-\frac4{13}x^{13})^{1/4}}$

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

$1=\dfrac{1^8}{(C-\frac4{13}1^{13})^{1/4}}$
$1=\dfrac1{(C-\frac4{13})^{1/4}}$
$1=\left(C-\dfrac4{13}\right)^{1/4}$
$1=C-\dfrac4{13}$
$C=\dfrac{17}{13}$

so that the particular solution is

$y=\dfrac{x^8}{(\frac{17}{13}-\frac4{13}x^{13})^{1/4}}$