Answer to Question #104800 in Differential Equations for Suraj Singh

Question

Solve the differential equation:
xdy - (3y + x^5. y^1/3)dx = 0

Answer 1

$xdy-(3y+x^5y^{\frac{1}{3}})dx=0\\ xy'-3y=x^5y^{\frac{1}{3}}$

Bernoulli equation

divisible by $y^{\frac{1}{3}}$

$\frac{xy'}{y^{\frac{1}{3}}}-3y^\frac{2}{3}=x^5$

we will replace

$y^{\frac{2}{3}}=z(x)\\ y=z^{\frac{3}{2}}\\ y'=\frac{3}{2}z^{\frac{1}{2}}z'\\ y^{\frac{1}{3}}=z^{\frac{1}{2}}$

then

$\frac{x\cdot\frac{3}{2}\cdot z^{\frac{1}{2}}\cdot z'}{z^{\frac{1}{2}}}-3z=x^5\\ \frac{3}{2}xz'-3z=x^5\\ 3xz'-6z=2x^5$

obtain a linear equation

1)

$3xz'-6z=0\\ \frac{dz}{z}=\frac{2dx}{x}\\ ln|z|=2ln|x|+ln|C|\\ z=C\cdot x^2$

2)

$z=C(x)\cdot x^2\\ z'=C'x^2+2Cx\\$

substitute into the equation

$3xz'-6z=2x^5$

$3x(C'x^2+2Cx)-6Cx^2=2x^5\\ 3x^3C'-6x^2C-6Cx^2=2x^5\\ C'=\frac{2}{3}x^2\\ C(x)=\frac{2}{3}\cdot\frac{x^3}{3}+C_1$

Then

$z=(\frac{2x^3}{9}+C_1)x^2\\ y^{\frac{2}{3}}=(\frac{2x^3}{9}+C_1)x^2\\$ is the general solution,

 $y=0$ is a solution