Question #10b00

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Question #10b00

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Answer 1 · 2017-03-28T06:28:43

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

Explanation:

given ylnydx+(x-lny) dy=0

$\Rightarrow \frac{d x}{d y} = \frac{ln y - x}{y ln y}$ $rArr(dx)/(dy)=(lny-x)/(ylny)$
$\frac{d x}{d y} = \frac{1}{y} - \frac{x}{y ln y}$ $(dx)/(dy)=1/y -x/(ylny)$
$\frac{d x}{d y} + (\frac{1}{y ln y}) x = \frac{1}{y}$ $(dx)/(dy) +(1/(ylny))x=1/y$ ......(1)

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

multiply equation (1) by $e^{\int P d y}$ $e^(intPdy)$ , we get
$e^{\int P d y} \frac{d x}{d y} + P x e^{\int P d y} = Q e^{\int P d y}$ $e^(intPdy)(dx)/(dy) +Pxe^(intPdy)=Qe^(intPdy)$
$\Rightarrow d \frac{x e^{\int P d y}}{d y} = Q e^{\int P d y}$ $rArr d(xe^(intPdy))/(dy)=Qe^(intPdy)$
$\Rightarrow d (x e^{\int P d y}) = Q e^{\int P d y} d y$ $rArrd(xe^(intPdy))=Qe^(intPdy)dy$

integrating both sides
$\int d (x e^{\int P d y}) = \int Q e^{\int P d y} d y$ $intd(xe^(intPdy))=intQe^(intPdy)dy$
$x e^{\int P d y} = \int Q e^{\int P d y} d y$ $xe^(intPdy)=intQe^(intPdy)dy$

substitute P and Q in the above equation

$x e^{\int (\frac{1}{y ln y}) d y} = \int \frac{1}{y} e^{\int (\frac{1}{y ln y}) d y} d y + c .$ $xe^(int(1/(ylny))dy)=int1/ye^(int(1/(ylny))dy)dy+c.$ ...(2)

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

so from (2)

$x ln y = \int \frac{1}{y} ln y d y + c .$ $xlny=int1/ylny dy+c.$

to solve R.H.S, using same substitution, you get $\int \frac{1}{y} ln y d y = \frac{{(ln y)}^{2}}{2}$ $int1/ylnydy=((lny)^2)/2$
$\Rightarrow x ln y = \frac{{(ln y)}^{2}}{2} + c .$ $rArr xlny=((lny)^2)/2+c.$
$\Rightarrow 2 x ln y = {(ln y)}^{2} + k, k = 2 c .$ $rArr2xlny=(lny)^2+k, k=2c.$
$\Rightarrow {(ln y)}^{2} - 2 x ln y + k = 0$ $rArr (lny)^2-2xlny+k=0$

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

$:. lny=x+-sqrt(x^2-k).$

$:. y=e^(x+-sqrt(x^2-k))," is the Gen. Soln."$

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

Answer 2 · 2017-03-28T10:36:37

$y = e^(xpm sqrt(x^2+C_1))$

Explanation:

$y log y dx+(x-log y)dy =0$

or

$(dy)/(dx)=-(y log y)/(x-log y)$ now making the transformation

$z=x-log y$ we have

$(dz)/(dx)=1-1/y(dy)/(dx)$ or

$(dy)/(dx)=y(1-(dz)/(dx))$

but

$y = e^(x-z)$ so putting all together

$e^(x-z)(1-(dz)/(dx))=-e^(x-z)((x-z)/z)$

considering that $e^(x-z) ne 0$

$1-(dz)/(dx)= -((x-z)/z)=1-x/z$ or

$(dz)/(dx)=x/z$

Now this differential equation is separable so

$z dz=x dx$ with solution

$z^2=x^2+C_1$

or

$(x-log y)^2=x^2+C_1$ then

$x-log y = pm sqrt(x^2+C_1)$ and finally

$y = e^(xpm sqrt(x^2+C_1))$

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