# Question 10b00

Mar 28, 2017

$y = {e}^{x \pm \sqrt{{x}^{2} - k}} , \text{ is the Gen. Soln.}$

#### Explanation:

given ylnydx+(x-lny) dy=0

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}} = \frac{\ln y - x}{y \ln y}$
$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{y} - \frac{x}{y \ln y}$
$\frac{\mathrm{dx}}{\mathrm{dy}} + \left(\frac{1}{y \ln y}\right) x = \frac{1}{y}$ ......(1)

this is a linear differential equation of the form $\left(\frac{\mathrm{dx}}{\mathrm{dy}} + P \cdot x = Q\right)$ , where P and Q are either constant or a function of 'y'
So integrating factor is ${e}^{\int P \mathrm{dy}}$, and here P = $\left(\frac{1}{y \ln y}\right)$ , Q =$\frac{1}{y}$

multiply equation (1) by ${e}^{\int P \mathrm{dy}}$, we get
${e}^{\int P \mathrm{dy}} \frac{\mathrm{dx}}{\mathrm{dy}} + P x {e}^{\int P \mathrm{dy}} = Q {e}^{\int P \mathrm{dy}}$
$\Rightarrow d \frac{x {e}^{\int P \mathrm{dy}}}{\mathrm{dy}} = Q {e}^{\int P \mathrm{dy}}$
$\Rightarrow d \left(x {e}^{\int P \mathrm{dy}}\right) = Q {e}^{\int P \mathrm{dy}} \mathrm{dy}$

integrating both sides
$\int d \left(x {e}^{\int P \mathrm{dy}}\right) = \int Q {e}^{\int P \mathrm{dy}} \mathrm{dy}$
$x {e}^{\int P \mathrm{dy}} = \int Q {e}^{\int P \mathrm{dy}} \mathrm{dy}$

substitute P and Q in the above equation

$x {e}^{\int \left(\frac{1}{y \ln y}\right) \mathrm{dy}} = \int \frac{1}{y} {e}^{\int \left(\frac{1}{y \ln y}\right) \mathrm{dy}} \mathrm{dy} + c .$ ...(2)

to solve the integrating factor, put lny=t$\Rightarrow \left(\frac{1}{y}\right) \mathrm{dy} = \mathrm{dt}$
rArr e^(int(1/(ylny))dy)=e^(int(1/tdt)$= {e}^{\ln} t = t = \ln y$

so from (2)

$x \ln y = \int \frac{1}{y} \ln y \mathrm{dy} + c .$

to solve R.H.S, using same substitution, you get $\int \frac{1}{y} \ln y \mathrm{dy} = \frac{{\left(\ln y\right)}^{2}}{2}$
$\Rightarrow x \ln y = \frac{{\left(\ln y\right)}^{2}}{2} + c .$
$\Rightarrow 2 x \ln y = {\left(\ln y\right)}^{2} + k , k = 2 c .$
$\Rightarrow {\left(\ln y\right)}^{2} - 2 x \ln y + k = 0$

rArr lny={2x+-sqrt(4x^2-4k)}/2...[because," the quadr. forml.]"

$\therefore \ln y = x \pm \sqrt{{x}^{2} - k} .$

$\therefore y = {e}^{x \pm \sqrt{{x}^{2} - k}} , \text{ is the Gen. Soln.}$

for particular solution, assuming constant c = 0,we get
y=e^(x+-sqrt(x^2-0)) rArry=e^(2x) rArr2x=lny"#

Mar 28, 2017

$y = {e}^{x \pm \sqrt{{x}^{2} + {C}_{1}}}$

#### Explanation:

$y \log y \mathrm{dx} + \left(x - \log y\right) \mathrm{dy} = 0$

or

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \log y}{x - \log y}$ now making the transformation

$z = x - \log y$ we have

$\frac{\mathrm{dz}}{\mathrm{dx}} = 1 - \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(1 - \frac{\mathrm{dz}}{\mathrm{dx}}\right)$

but

$y = {e}^{x - z}$ so putting all together

${e}^{x - z} \left(1 - \frac{\mathrm{dz}}{\mathrm{dx}}\right) = - {e}^{x - z} \left(\frac{x - z}{z}\right)$

considering that ${e}^{x - z} \ne 0$

$1 - \frac{\mathrm{dz}}{\mathrm{dx}} = - \left(\frac{x - z}{z}\right) = 1 - \frac{x}{z}$ or

$\frac{\mathrm{dz}}{\mathrm{dx}} = \frac{x}{z}$

Now this differential equation is separable so

$z \mathrm{dz} = x \mathrm{dx}$ with solution

${z}^{2} = {x}^{2} + {C}_{1}$

or

${\left(x - \log y\right)}^{2} = {x}^{2} + {C}_{1}$ then

$x - \log y = \pm \sqrt{{x}^{2} + {C}_{1}}$ and finally

$y = {e}^{x \pm \sqrt{{x}^{2} + {C}_{1}}}$