solving differential equation using Lagrange method

[Elena Baby]

Here is the problem:



I divided both sides by: Sin^2(X) and here is what I got:



I found the other public answer of the homogenous part which equals to y₂=x

From what I know, I should solve the private answer of the non-homogenous part using the Lagrange method.

So:Y_g = v₁y₁+v₂y₂.

I found out that v₂ equals x,but when I tried finding v₁, here is what I got:



Do you think I made any mistakes? If not, how can I solve this integral? Thank you for helping!

 

[Elena Baby]

Oh, I think I know where the problem is!y₂ should equal to vy₁ but I only took v as y₂.

 

[HallsofIvy]

Here is the problem:

View attachment 19591

I divided both sides by: Sin^2(X) and here is what I got:

View attachment 19590

I found the other public answer of the homogenous part which equals to y₂=x

?? If y= x then y'= 1 and y''= 0 so y''- 2 cot(x)y'+ y= -2cot(x)+ x, not 0.

From what I know, I should solve the private answer of the non-homogenous part using the Lagrange method.

So:Y_g = v₁y₁+v₂y₂.

I found out that v₂ equals x, but when I tried finding v₁, here is what I got:

View attachment 19592

Do you think I made any mistakes? If not, how can I solve this integral? Thank you for helping!

 

[Steven G]

I think that sin^2(x) and cos^2(x) + 1 are not equal! So (cos^2(x) + 1)y/sin^2(x) is not y. What do you think??

 

[Elena Baby]

I think that sin^2(x) and cos^2(x) + 1 are not equal! So (cos^2(x) + 1)y/sin^2(x) is not y. What do you think??

Oh My God!How did I even-Thank you for pointing out!