Question

How to I solve homogeneous equation of `y dy/dx +x=2y` ?

Answer 1

The general solution is `ln(|y/x-1|)-x/(y-1)=-ln(|x|)+C`

Explanation:

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`

Answer 2

Substitute `u=y/x` and solve

Explanation:

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.

Answer 3

The general solution of given equation is :
`y=x+c*e^(x/(x-y)`

Explanation:

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.