Question

Question #a1a8c

Answer 1

`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.

Answer 2

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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