Given the differential Equation (dy/dx)+(2/x)y=x^2y^3 ;solve this equation using the Bernoulli method; Final answer should be (1/y^2)=?

Answer:[tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].Step-by-step explanation:Given differential equation[tex]\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{2}{x}y=x^2y^3[/tex]Differential equation can be write as [tex]y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{2}{x}y^{-2}=x^2[/tex]By Bernoulli methodSusbstitute [tex]y^{-2}=t[/tex].....{equationI}Differentiate equation I w.r.t x then we get [tex]\frac{\mathrm{d}t}{\mathrm{d}x}=-2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}[/tex][tex]-\frac{1}{2}\frac{\mathrm{d}t}{\mathrm{d}x}=y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}[/tex]Susbstitute the values in the given differential equation then we get [tex]-\frac{1}{2}\frac{\mathrm{d}t}{\mathrm{d}x}+\frac{2}{x}t=x^2[/tex][tex]\frac{\mathrm{d}t}{\mathrm{d}x}-\frac{4}{x}t=-2x^2[/tex]It is first order linear differential equation and compare with the first order linear differential equation [tex]\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)[/tex]Then we get P(x)=[tex]-\frac{4}{x}[/tex] and Q(x)=[tex]-2x^2[/tex]Integration factor=[tex]e^\intP(x)dx[/tex]Integration factor= [tex]e^{-\int\frac{4}{x}dx[/tex]Integration factor= [tex]e^{-4lnx}=e^{lnx^{-4}}=x^{-4}[/tex].Using [tex]e^{logb}=b[/tex][tex]t\times \frac{1}{x^4}=\int{-2x^2}\times\frac{1}{x^4}dx+C[/tex][tex]t=-2x^4{\intx^{-2}dx+C}[/tex][tex]t=2x^4\times\frac{1}{x}+Cx^4[/tex][tex]t=2x^3+Cx^4[/tex]Substitute [tex]t=\frac{1}{y^2}[/tex] then we get [tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].Answer: [tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].