2.12 Example 12 (y)2x+yylnyy2(lny)4=0

(y)2x+yylnyy2(lny)4=0p2x+pylnyy2(lny)4=0

Applying p-discriminant method gives

F=p2x+pylnyy2(lny)4=0Fp=2px+ylny=0

We first check that Fy0.  Now we apply p-discriminant. Eliminating p. Second equation gives p=y2xlny. Substituting into the first equation gives

(y2xlny)2x+(y2xlny)ylnyy2(lny)4=0y24x(lny)2y22x(lny)2y2(lny)4=0y2ln(y)2(14x12x(lny)2)=0

Hence we obtain the solutions

y=0y=114x12x(lny)2=0

Or

(1)y=0(2)y=1(3)1+4x(lny)2=0

The solution y=0 does not satisfy the ode. But y=1 does. The solution 1+4x(lny)2=0 gives y={ei2xei2x But these do not satisfy the ode.

The primitive can be found to be

Ψ(x,y,c)=yecc2x=0

Now we have to eliminate c using the c-discriminant method

Ψ(x,y,c)=yecc2x=0Ψ(x,y,c)c=(1c2x2c2(c2x)2)ecc2x=0

Second equation gives 1c2x2c2(c2x)2=0 or ecc2x=0. For the first one, c=±x. Substituting x in first equation gives

yexxx=0y=ex2xlny=x2x(lny)2=x4x24x(lny)2+1=0(4)ys={ei2xei2x

These do not satisfy the ode. Can not obtain y=1 solution using c-discriminant.

See paper by C.N. SRINIVASINGAR, example 4. The following plot shows the singular solution above as the envelope of the family of general solution plotted using different values of c. Added also ys=1.