Answer:
[tex]y=\displaystyle\frac{2\ln x}{x}+\frac{1}{x}[/tex]
Step-by-step explanation:
Divide both sides of the equation by [tex]x^2[/tex]:
[tex]\displaystyle y'-\frac{1}{x}y=\frac{2}{x^2}[/tex]
Find the integrating factor:
[tex]u=e^{\int\frac{1}{x}}=e^{\ln x}=x[/tex]
Multiply the equation by the integrating factor:
[tex]\displaystyle xy'-y=\frac{2}{x}[/tex]
The left side is the derivative of (xy), therefore we can write the equation as:
[tex]\displaystyle\frac{d(xy)}{dx}=\frac{2}{x}[/tex]
That is a separable equation. We separate and integrate:
[tex]\displaystyle\int d(xy)=\int\frac{2}{x}dx[/tex]
We get:
[tex]xy=2\ln x + C[/tex]
Then plug the initial value y(1)=1, which means to plug x=1 and y=1:
[tex]1(1)=2\ln 1 + C\to 1= C[/tex]
Therefore, the solution once we plug C=1 becomes:
[tex]xy=2\ln x+1[/tex]
Then solving for y by dividing both sides by x, we get:
[tex]y=\displaystyle\frac{2\ln x}{x}+\frac{1}{x}[/tex]
