Question #a1a8c

2 Answers
Nov 27, 2016

#Pi_(k=1)^3(y/x+r_k)^(a_k)=C_1/x#

Explanation:

Making #y(x) = lambda(x) x# and substituting into the differential equation we obtain

#(dlambda)/(dx)=-(lambda^3+6lambda+5)/(xlambda(lambda+5))#. This differential equation is separable so

#((lambda^2+5lambda)dlambda)/(lambda^3+6lambda+5)=-dx/x#

This equation can be expanded as

#(a_1/(lambda+r_1)+a_2/(lambda+r_2)+a_3/(lambda+r_3))dlambda=-dx/x#

where #r_1,r_2,r_3# are the roots of #lambda^3+6lambda+5=0#

After integrating we have

#sum_(k=1)^3a_klogabs (lambda+r_k)=-logabsx+C#

or

#Pi_k(lambda+r_k)^(a_k)=C_1/x#

or

#Pi_(k=1)^3(y/x+r_k)^(a_k)=C_1/x#

We should consider also the solution #y=0# eliminated in the substitution process.

Nov 28, 2016

See explanation

Explanation:

This is not an exact differential equation. As the coefficients of dx and dy

are homogeneous functions of x and y, the substitution y = vx would

help solving this differential equation.

Eliminating y,

#v + x (dv)/(dx)=(dy)/(dx)=-(5+v^2)/(5v+v^2)#

Separating variables and integrating,

#int (dx)/x = -int 1/(v+(5v+v^2)/(5+v^2)) dv#

#ln x =- int (5+v^2)/(v(5v+v^2)+5+v^2) dv#

#=- int (5+v^2)/(v^3+6v^2+5) dv#

Note that there is only one ( negative ) real zero #v_1# for #v^3+6v^2+5#

Assume that #v^3+6v^2+5 = (v-v_1)((v-alpha)^2+beta^2)# and

resolve the integrand into partial fractions, in the form

#P/(v-v_1)+(Q(v-alpha)+R)/((v-alpha)^2+beta^2)#,

where, like #v_1 , alpha and beta#,

P, Q and R are known constants. Now, the solution is

#ln x =-ln(v-v_1)^P-int (Q(v-alpha)+R)/((v-alpha)^2+beta^2) dv#.

Upon integration and rearrangement,

#ln x (v-v_1)^P((v-alpha)^2+beta^2)^(Q/2) + R/beta tan^(-1)((v-alpha)/beta)+ C#

Reverting to y,

#x(y/x-v_1)^P((y/x-alpha)^2+beta^2)^(Q/2)+R/beta tan^(-1)((y/x-alpha)/beta))=C#

It is for the interested reader to evaluate the constants

#v_1, alpha, beta, P, Q and R#.

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