Respuesta :
Answer:
Therefore , every element of the function [tex]y =\frac{6 \ ln (x)+c}{x}[/tex] is a solution of the DE [tex]x^2y'+xy=6[/tex].
Step-by-step explanation:
Given differential equation is
[tex]x^2y'+xy=6[/tex]
[tex]\Rightarrow y'+\frac{xy}{x^2}=\frac{6}{x^2}[/tex] [ divided by x²]
[tex]\Rightarrow y'+\frac{y}{x}=\frac{6}{x^2}[/tex]
Here [tex]P= \frac{1}{x}[/tex] and [tex]Q=\frac{6}{x^2}[/tex]
The integrating factor is = [tex]e^{\int pdx}[/tex]
[tex]=e^{\int \frac{1}{x}dx[/tex]
[tex]=e^{ln x}[/tex]
= x
Multiplying the integrating factor both sides of the ODE
[tex]x y'+x.\frac{y}{x}=x. \frac{6}{x^2}[/tex]
[tex]\Rightarrow xy'+y=\frac{6}{x}[/tex]
[tex]\Rightarrow x\frac{dy}{dx}+y=\frac{6}{x}[/tex] [ [tex]\because y'=\frac{dy}{dx}[/tex] ]
[tex]\Rightarrow xdy+ydx=\frac{6}{x}dx[/tex]
Integrating both sides
[tex]\int xdy+\int ydx=\int\frac{6}{x}dx[/tex]
[tex]\Rightarrow xy= 6\ ln(x)+c[/tex] [ c is an arbitrary constant]
[tex]\therefore y =\frac{6 \ ln (x)+c}{x}[/tex] for x>0
Therefore , every element of the function [tex]y =\frac{6 \ ln (x)+c}{x}[/tex] is a solution of the DE [tex]x^2y'+xy=6[/tex].
