Question #10b00

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"#

#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))#