Answer to Question #105507 in Differential Equations for khushi

Question

Identify the type of the differential equation y =xy'+1-lny' and hence
solve it

Answer 1

$y=-ln(dy/dx)+xdy/dx+1$

This is Clairaut equation

Differentiate both sides with respect to x

${\frac {dy} {dx}}=x{\frac {d^2y} {dx^2}}+{\frac {dy} {dx}}-{\frac {{\frac {d^2y} {dx^2}}} {{\frac {dy} {dx}}}}$

Collect in terms of ${\frac {d^2y} {dx^2}}$

${\frac {dy} {dx}}={\frac {dy} {dx}}+{\frac {d^2y} {dx^2}}(x-{\frac {1} {{\frac {dy} {dx}}}})$

Then

${\frac {d^2y} {dx^2}}(x-{\frac {1} {{\frac {dy} {dx}}}})=0$

Solve

${\frac {d^2y} {dx^2}}=0$ and $(x-{\frac {1} {{\frac {dy} {dx}}}})=0$ separately:

For ${\frac {d^2y} {dx^2}}=0$ :

${\frac {dy} {dx}}=\int 0dx=C_1$

Substitute $y'=C_1$ into initial equation

$y=-ln(C_1)+C_1x+1$

For $(x-{\frac {1} {{\frac {dy} {dx}}}})=0$ :

${\frac {dy} {dx}}={\frac {1} {x}}$

Substitute $y'=1/x$ into initial equation

$y=-ln({\frac {1} {x}})+2$

Answer: $y=-ln(C_1)+C_1x+1$ or $y=-ln({\frac {1} {x}})+2$