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

this is a problem of variation of parameter

2 Answers
Dec 16, 2017

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))

Dec 17, 2017

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