# Identify the following differential equations and hence solve it y=xy'+1-ln y'?

Jul 3, 2018

General Solution:

• $\boldsymbol{y = c x + 1 - \ln c}$

and singular solution:

• $y = 2 + \ln x$

#### Explanation:

This is in the form of Clairaut's equation:

$y = y ' x + {\underbrace{F \left(y '\right)}}_{= 1 - \ln y '} q \quad \square$

Differentiate wrt $x$:

$\cancel{y '} = \cancel{y '} + y ' ' x + \frac{\mathrm{dF}}{d y '} y ' '$

And factor:

$\therefore y ' ' \left(x + \frac{\mathrm{dF}}{d y '}\right) = 0 \implies \left\{\begin{matrix}y ' ' = 0 q \quad q \quad \boldsymbol{X} \\ \frac{\mathrm{dF}}{d y '} = - x q \quad \boldsymbol{Y}\end{matrix}\right.$

• $\boldsymbol{X}$

$y ' ' = 0 \implies \left\{\begin{matrix}y ' = {c}_{1} \\ y = {c}_{1} x + {c}_{2}\end{matrix}\right. q \quad \star$

Subbing both $\star$ equations into $\square$:

${c}_{1} x + {c}_{2} = {c}_{1} x + F \left({c}_{1}\right) \implies {c}_{2} = F \left({c}_{1}\right)$

This gives the general solution:

$\boldsymbol{y = c x + 1 - \ln c}$

• $\boldsymbol{Y}$

$\frac{d}{\mathrm{dy} '} \left(1 - \ln y '\right) = - x q \quad \implies - \frac{1}{y '} = - x$

$\implies y ' = \frac{1}{x}$

Putting that into $\square$:

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

$y = 2 + \ln x$

This is a singular solution.