Question

The solution to the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition y(1) = 1 is-

• A. y = x lnx + x^2
• B. y = xe^(x–1)
• C. y = x lnx + x
• D. y = lnx + x

Answer 1

Answer: (C)

y = x lnx + x

Answer 2

$\frac{dy}{dx}$= 1 + $\frac{y}{x}$ $\frac{y}{x}$= v

v + x $\frac{dv}{dx}$= 1 + v Ž v = log x + c by y(1) = 1 Ž $\frac{y}{x}$= log

x + c Ž 1 = c so $\frac{y}{x}$ = log x + 1