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 method
Susbstitute [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].
