Question

`4y''+36y="cosec"(3x)`, can u please help me to solve this equation?

this is a problem of variation of parameter

Answer 1

See below.

Explanation:

This is a linear non-homogeneous differential equation. Those DE heve a solution that can be composed as

`y = y_h + y_p` with

`4y_h''+36y_h=0` and

`4y_p''+36y_p="cosec"(3x)`

The solution for

`y_h''+9y_h=0` is easily determined

(see my answer to https://socratic.org/questions/y-9y-0-y1-sin3x-by-using-reduction-of-order-method-how-can-i-solve-this-equation )

Now the particular solution for

`4y_p''+36y_p="cosec"(3x)`

is `y_p = 1/36 ( Log_e(Sin(3 x)) Sin(3 x)-3 x Cos(3 x))`

and finally

`y = C_1 sin(3x)+C_2 cos(3x) + 1/36 ( Log_e(Sin(3 x)) Sin(3 x)-3 x Cos(3 x))`

Answer 2

`y=c_1sin3x+c_2cos3x+1/36Ln(sin3x)*sin3x-x/12*cos3x`

Explanation:

`4y''+36y=csc3x`

`y''+9y=1/4csc3x`

Characteristic equation of this differential one: `r^2+9=0`

Its roots are `r_1=-3i` and `r_2=3i`

Consequently homogeneous solution of it is:

`y_h=c_1sin3x+c_2cos3x`

I used variation of parameters for finding particular solution

`y_p=u_1*y_1+u_2*y_2`

In this case `y_1=sin3x` and `y_2=cos3x`

According to this method, `u_1` and `u_2` had to 2 conditions:

`(u_1)'*sin3x+(u_2)'*cos3x=0` and `3(u_1)'*cos3x-3(u_2)'*sin3x=1/4csc3x`

From first equation, `(u_2)'=-(u_1)'*tan3x`

Hence,

`3(u_1)'cos3x-3*(-(u_1)'tan3x)sin3x=1/4csc3x`

`3(u_1)'*cos3x+3(u_1)'*(sin3x)^2/(cos3x)=1/4csc3x`

`3(u_1)'*(cos3x)^2+3(u_1)'*(sin3x)^2=1/4csc3x*cos3x`

`3(u_1)'*[(cos3x)^2+(sin3x)^2]=1/4cot3x`

`3(u_1)'=1/4cot3x`

`(u_1)'=1/12cot3x`

`u_1=1/36ln(sin3x)`

Consequently,

`(u_2)'=-(u_1)'*tan3x`

`(u_2)'=-1/12cot3x*tan3x`

`(u_2)'=-1/12`

`u_2=-x/12`

Thus,

`y_p=1/36Ln(sin3x)*sin3x-x/12*cos3x`

`y=y_h+y_p=c_1sin3x+c_2cos3x+1/36Ln(sin3x)*sin3x-x/12*cos3x`