Question

Find the general solution of the Bernoulli differential equation: y' + 4y = xy^4. Use lower case c for the constant in answer.

Answer 1

Answer with Step-by-step explanation:

The given equation is

$\frac{dy}{dx}+4y=xy^4$

Dividing throughout by $y^4$ we get

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

Substituting $t=y^{-3}$ in the above equation we get

$dt=-3y^{-4}\cdot \frac{dy}{dx}\\\\y^{-4}\cdot {dy}=\frac{-dt}{3}$

Thus we get

$\frac{-dt}{dx}+12t=3x$

Which is a linear differential equation of the form

$t'+p(x)t=q(x)dx}$[/tex]

whose solution is given by

$te^{\int p(x)dx}=\int (e^{\int p(x)}q(x)dx$

Solving we get

$te^{12x}=\int 3e^{12x}\times xdx\\\\te^{12x}=3x\cdot \frac{e^{12x}}{12}-\int 3\times \frac{e^{12x}}{12}dx\\\\te^{12x}=\frac{xe^{12x}}{4}-\frac{e^{12x}}{48}\\\\t=\frac{x}{4}-\frac{1}{48}\\\\\frac{1}{y^3}=\frac{x}{4}-\frac{1}{48}+c$