How to I solve homogeneous equation of #y dy/dx +x=2y# ?
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The general solution is #ln(|y/x-1|)-x/(y-1)=-ln(|x|)+C#
The ODE is
#ydy/dx+x=2y#
Divide by #y#
#dy/dx+x/y=2#
Let #y=vx#
Then,
#dy/dx=v+x(dv)/dx#
Substituting in the ODE
#v+x(dv)/dx+1/v=2#
#x(dv)/dx=2-v-1/v=(2v-v^2-1)/(v)=-(v^2-2v+1)/v#
Therefore,
#(vdv)/(v-1)^2=-dx/x#
Integrating both sides
#int(vdv)/(v-1)^2=-intdx/x=-lnx+C#
For the #LHS#,
Let #u=v-1#, #=>#, ##du=dv
#int(vdv)/(v-1)^2=int((u+1)du)/u^2#
#=int(1/u+1/u^2)du#
#=ln(u)-1/u#
#=ln(v-1)-1/(v-1)#
#=ln(y/x-1)-1/(y/x-1)#
#=ln(y/x-1)-x/(y-x)#
The general solution is
#ln(y/x-1)-x/(y-x)=-ln(x)+C#
Where #C in RR#
Substitute #u=y/x# and solve
Be careful with terminology here - "homogeneous ordinary differential equation" can mean two entirely different things!
1) An equation in #y# and its derivatives w.r.t. #x# where all coefficients are functions of #x# alone.
2) A first-order ODE where #dy/dx# is equal to a function of #y/x#. This type of ODE can be solved by the substitution #u=y/x#.
In this question we are dealing with the second meaning; let's rearrange into the needed form:
#ydy/dx+x=2y#
#dy/dx=2-x/y#
Substitute #u=y/x#. Note that by the quotient rule #(du)/dx=d/dx(y/x)=(xdy/dx-y)/x^2=1/xdy/dx-y/x^2=1/xdy/dx-u/x#
so
#dy/dx=x(du)/dx+u#
Substituting in:
#x(du)/dx+u=2-1/u#
#x(du)/dx=2-u-1/u#
#1/(2-u-1/u)(du)/dx=1/x#
This is now a separable equation, so integrate for the solution:
#int(du)/(2-u-1/u)=intdx/x#
The #x# integral is straightforward - the natural logarithm #ln|x|+C#. The #u# integral needs some rearrangement.
#int(du)/(2-u-1/u)=intu/(2u-u^2-1)du=int-u/(u^2-2u+1)du#
#=int-u/(u-1)^2du=int-(u-1+1)/(u-1)^2du#
#=int-(u-1)/(u-1)^2-1/(u-1)^2du#
#=int-1/(u-1)-1/(u-1)^2du#
#=-ln|u-1|+1/(u-1)+C#
Put the two integral solutions together:
#-ln|u-1|+1/(u-1)=ln|x|+C#
Now substitute back for #y#:
#-ln|y/x-1|+1/(y/x-1)=ln|x|+C#
Some tidying up rearrangement:
#-ln|(y-x)/x|+x/(y-x)=ln|x|+C#
#-ln|y-x|+ln|x|+x/(y-x)=ln|x|+C#
#-ln|y-x|+x/(y-x)=C#
#ln|y-x|-x/(y-x)=C#
As we have both #y# and #x# both inside and outside of the logarithm there's no beautiful way to express this as a function of one in terms of the other; this is as tidy as we get.
The general solution of given equation is :
#y=x+c*e^(x/(x-y)#
Here,
#y(dy)/(dx)+x=2y#
#=>y(dy)/(dx)=2y-x#
#=>(dy)/(dx)=(2y-x)/y#
#=>(dy)/(dx)=(2(y/x)-1)/(y/x)to#[homogeneous euqn.]
Subst. #y/x=v=>y=vx=>(dy)/(dx)=v+x(dv)/(dx)#
So,
#v+x(dv)/(dx)=(2v-1)/v#
#=>x(dv)/(dx)=(2v-1)/v-v=(2v-1-v^2)/v=-(v^2-2v+1)/v#
#=>x(dv)/(dx)=-(v-1)^2/v#
#=>v/(v-1)^2dv=-1/xdx#
Integrating both sides:
#intv/(v-1)^2dv=int-1/xdx#
#=>int((v-1)+1)/(v-1)^2dv=-int1/xdx#
#=>int[1/(v-1)+1/(v-1)^2]dv=-ln|x|+lnc#
#=>ln|v-1|-1/(v-1)=-ln|x|+lnc#
#=>ln|v-1|+ln|x|-lnc=1/(v-1)#
#ln|((v-1)x)/c|=-1/(v-1)#
Subst. back ,#y/x=v # ,we get
#ln|((y/x-1)x)/c|=-1/(y/x-1)#
#=>ln|(y-x)/c|=-x/(y-x)#
#=>ln|(y-x)/c|=x/(x-y)#
#(y-x)/c=e^(x/(x-y)#
#y-x=c*e^(x/(x-y)#
#=>y=x+c*e^(x/(x-y)#
This is the general solution of given equation.
