Question

Question #10b00

Answer 1

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

Explanation:

given ylnydx+(x-lny) dy=0

`rArr(dx)/(dy)=(lny-x)/(ylny)`
`(dx)/(dy)=1/y -x/(ylny)`
`(dx)/(dy) +(1/(ylny))x=1/y` ......(1)

this is a linear differential equation of the form `((dx)/(dy) +P*x=Q)` , where P and Q are either constant or a function of 'y'
So integrating factor is `e^(intPdy)`, and here P = `(1/(ylny))` , Q =`1/y`

multiply equation (1) by `e^(intPdy)`, we get
`e^(intPdy)(dx)/(dy) +Pxe^(intPdy)=Qe^(intPdy)`
`rArr d(xe^(intPdy))/(dy)=Qe^(intPdy)`
`rArrd(xe^(intPdy))=Qe^(intPdy)dy`

integrating both sides
`intd(xe^(intPdy))=intQe^(intPdy)dy`
`xe^(intPdy)=intQe^(intPdy)dy`

substitute P and Q in the above equation

`xe^(int(1/(ylny))dy)=int1/ye^(int(1/(ylny))dy)dy+c.` ...(2)

to solve the integrating factor, put lny=t`rArr (1/y)dy=dt`
`rArr e^(int(1/(ylny))dy)=e^(int(1/tdt)` `= e^lnt = t = lny`

so from (2)

`xlny=int1/ylny dy+c.`

to solve R.H.S, using same substitution, you get `int1/ylnydy=((lny)^2)/2`
`rArr xlny=((lny)^2)/2+c.`
`rArr2xlny=(lny)^2+k, k=2c.`
`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

`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))`