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

Answer with Step-by-step explanation:The given equation is [tex]\frac{dy}{dx}+4y=xy^4[/tex]Dividing throughout by [tex]y^4[/tex] we get[tex]\frac{dy}{y^4dx}+\frac{4}{y^3}=x\\\\y^{-4}\frac{dy}{dx}+4y^{-3}=x\\\\[/tex]Substituting [tex]t=y^{-3}[/tex] in the above equation we get[tex]dt=-3y^{-4}\cdot \frac{dy}{dx}\\\\y^{-4}\cdot {dy}=\frac{-dt}{3}[/tex]Thus we get[tex]\frac{-dt}{dx}+12t=3x[/tex]Which is a linear differential equation of the form[tex]t'+p(x)t=q(x)dx}[/tex][/tex]whose solution is given by[tex]te^{\int p(x)dx}=\int (e^{\int p(x)}q(x)dx[/tex]Solving we get[tex]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[/tex]