solve: [(d^3y)/(dx^3)]+[(dy)/(dx)]=Cosec x

Question

solve using variation of parameters

Answer 1

∫cosec(x)dx=∫(cot(x)+cosec(x))cosec(x)dx/(cot(x)+cosec(x)),

∫cosec(x)dx=-∫(-cot(x)-cosec(x))cosec(x)dx/(cot(x)+cosec(x)),

∫cosec(x)dx=-∫(-cot(x)cosec(x)-cosec²(x))dx(cosec(x)+cot(x)).

∫cosec(x)dx=-ln|cosec(x)+cot(x)|+C, where C is integration constant.

cosec(x)+cot(x)=1/sin(x)+cos(x)/sin(x)=(1+cos(x))/sin(x).

-ln|cosec(x)+cot(x)|=ln|1/(cosec(x)+cot(x))|=ln|sin(x)/(1+cos(x))|=ln|sin(x)|-ln|1+cos(x)|.

d³y/dx³+dy/dx=cosec(x)=

(d/dx)(d²y/dx²+y)=(d/dx)(∫cosec(x)dx), then integrating:

d²y/dx²+y=f(x), where f(x)=ln|sin(x)|-ln|1+cos(x)|+C.

d²y/dx²+y=0 has the solution y_c (characteristic): Asin(x)+Bcos(x), where A and B are constants.

Now we need the particular solution y_p.

y=y_c+y_p. Here we start using variation of parameters.

We already have y_c=Asin(x)+Bcos(x), now we use y₁=sin(x) and y₂=cos(x), where:

y_p=u₁sin(x)+u₂cos(x). u₁ and u₂ are functions of x and we find them using Wronskian determinants w, w_1, w₂, and Cramer's Rule for solving simultaneous equations.

w=

| sin(x)  cos(x) |

| cos(x) -sin(x) | = -sin²(x)-cos²(x)=-1.

We have here y₁ and y₂ in the first row and their respective derivatives in the second row.

Now Cramer's Rule introducing f(x). (See above for definition of f(x).)

w₁=

|   0   cos(x) |

| f(x) -sin(x) | = -f(x)cos(x).

w₂=

| sin(x)    0  |

| cos(x) f(x) | = f(x)sin(x).

u₁'=w₁/w=f(x)cos(x); u₂'=w₂/w=-f(x)sin(x). Integrate to find u₁ and u₂:

u₁=∫(ln|sin(x)|-ln|1+cos(x)|+C)cos(x)dx=∫(ln|sin(x)|cos(x)dx-∫ln|1+cos(x)|cos(x)dx+C∫cos(x)dx;

u₂=∫(-ln|sin(x)|+ln|1+cos(x)|-C)sin(x)dx=∫(-ln|sin(x)|sin(x)dx+∫ln|1+cos(x)|sin(x)dx-C∫sin(x)dx.

Consider the integrals separately: u₁=I+J+K, u₂=L+M+N.

I=∫(ln|sin(x)|cos(x)dx. Let u=ln|sin(x)|, then e^u=sin(x), e^udu=cos(x)dx.

I=∫ue^udu; let p=u, then dp=du; let dq=e^udu, then q=e^u (integration by parts).

I=pq-∫qdp, that is:

I=ue^u-∫e^udu=ue^u-e^u=sin(x)ln|sin(x)|-sin(x).

K=C∫cos(x)dx=Csin(x); N=-C∫sin(x)dx=Ccos(x).

M=∫ln|1+cos(x)|sin(x)dx. Let u=ln|1+cos(x)|, then e^u=1+cos(x), e^udu=-sin(x)dx.

So M=-ue^u+e^u=-(1+cos(x))ln|1+cos(x)|+1+cos(x).

J=-∫ln|1+cos(x)|cos(x)dx and L=∫(ln|sin(x)|(-sin(x))dx remain to be integrated.

For J, let u=ln|1+cos(x)|, then du=-sin(x)dx/(1+cos(x)); let dv=cos(x)dx, then v=sin(x).

J=-sin(x)ln|1+cos(x)|-∫sin²(x)dx/(1+cos(x))=

-sin(x)ln|1+cos(x)|-∫(1-cos²(x))dx/(1+cos(x))=

-sin(x)ln|1+cos(x)|-∫(1-cos(x)dx=

-sin(x)ln|1+cos(x)|-x+sin(x).

For L, let u=ln|sin(x)|, then du=cot(x)dx; let dv=-sin(x)dx, then v=cos(x).

L=uv-∫vdu, that is:

L=cos(x)ln|sin(x)|-∫cos²(x)dx/sin(x)=cos(x)ln|sin(x)|-∫(1-sin²(x))dx/sin(x),

L=cos(x)ln|sin(x)|-∫(cosec(x)-sin(x))dx,

L=cos(x)ln|sin(x)|+ln|cosec(x)+cot(x)|-cos(x)=cos(x)ln|sin(x)|+ln|1+cos(x)|-ln|sin(x)|-cos(x).

u₁=I+J+K, u₂=L+M+N.

u₁=sin(x)ln|sin(x)|-sin(x)-sin(x)ln|1+cos(x)|-x+sin(x)+Csin(x),

u₁=sin(x)ln|sin(x)|-sin(x)ln|1+cos(x)|-x+Csin(x).

u₂=cos(x)ln|sin(x)|+ln|1+cos(x)|-ln|sin(x)|-cos(x)-(1+cos(x))ln|1+cos(x)|+1+cos(x)+Ccos(x),

u₂=cos(x)ln|sin(x)|-ln|sin(x)|-cos(x)ln|1+cos(x)|+1+Ccos(x).

y_p=u₁sin(x)+u₂cos(x)=

sin²(x)ln|sin(x)|-sin²(x)ln|1+cos(x)|-xsin(x)+Csin²(x)+

cos²(x)ln|sin(x)|-cos²(x)ln|1+cos(x)|-cos(x)ln|sin(x)|+cos(x)+Ccos²(x)=

ln|sin(x)|-ln|1+cos(x)|-xsin(x)-cos(x)ln|sin(x)|+cos(x)+C.

The full solution is:

y=Asin(x)+Bcos(x)+ln|sin(x)|-ln|1+cos(x)|-xsin(x)-cos(x)ln|sin(x)|+C. (A cos(x) term has been absorbed into the constant B.)