Q:

Differential Equations: Solve the initial value problemx2y' - xy = 2y(1) = 1

Accepted Solution

A:
Answer: Β [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]