Finding adjoint operator of inhomogenous ODE: L[y] = x2 y''+xy' - 4y = 3x (x>0)

[Mathsgla]

So I have this question that I'm stuck with. The question is:

Solve the following ODE by using an Integrating factor: L[y] = x² y''+xy' - 4y = 3x (x>0) and find the corresponding adjoint operator.

So my method was to multiply the ODE by an integrating factor z(x) and rewrite the ODE as follows:

zx²y''+zxy' -4zy = 3xz = Uy'' + (U'+V)y' + V'y

Then compare coefficients of y'',y',y to get

U = zx2 (1)
U'+V = xz (2)
V' = -4z (3)

Then I did U''+v' by differentiating (1) twice and adding (3) and differentiated (2) to get an equal value.

So my adjoint equation after computing this is L_adjoint[z]= x²z'' +3xz' -3.

So I don't think this is the correct answer for the adjoint. Also, I followed this through trying solutions of L_adjoint[z] = 0 in the form of z=x^m and getting a value of m to be 1 which doesn't solve the adjoint equal to zero.

Not quite sure how to proceed from here. Any help is greatly appreciated!

 

[HallsofIvy]

So I have this question that I'm stuck with. The question is:

Solve the following ODE by using an Integrating factor: L[y] = x² y''+xy' - 4y = 3x (x>0) and find the corresponding adjoint operator.

So my method was to multiply the ODE by an integrating factor z(x) and rewrite the ODE as follows:

zx²y''+zxy' -4zy = 3xz = Uy'' + (U'+V)y' + V'y

Then compare coefficients of y'',y',y to get

U = zx2 (1)
U'+V = xz (2)
V' = -4z (3)

Then I did U''+v' by differentiating (1) twice and adding (3) and differentiated (2) to get an equal value.

So my adjoint equation after computing this is L_adjoint[z]= x²z'' +3xz' -3.

First, this is not an equation. I presume you mean that the this is the "adjoint operator" (and actually it is x²z''+ 3xz'- 3z) and that the adjoint equation is x²z''+ 3xz'- 3z= 0.

So I don't think this is the correct answer for the adjoint. Also, I followed this through trying solutions of L_adjoint[z] = 0 in the form of z=x^m and getting a value of m to be 1 which doesn't solve the adjoint equal to zero.

Not quite sure how to proceed from here. Any help is greatly appreciated!

Trying z= x^m, z'= mx^m-1, and z''= m(m-1)x^m-2. Putting those into the equation, x²(m(m- 1)x^{m- 2})+ 3x(mx^m-1)- 3x^m= (m(m- 1)- 3m+ 3)x^m= 0. Since x can be non-zero, in order that this not be identically 0 we must have m(m- 1)- 3m+ 3= m²- 4m+ 3= (m- 1)(m- 3)= 0. Yes, m= 1 and m= 3 are roots which means that two independent solutions to the associated homogeneous equation are x and x³. I don't know what you mean by "which doesn't solve the adjoint equal to 0". m= 2 give z= x so that z'= 1 and z''= 0. utting those into the equation, x²(0)+ 3x(1)- 3x= 0 is certainly true. The general solution to the associated homogeneous equation is z= Ax+ Bx³.

 

[Mathsgla]

First, this is not an equation. I presume you mean that the this is the "adjoint operator" (and actually it is x²z''+ 3xz'- 3z) and that the adjoint equation is x²z''+ 3xz'- 3z= 0.


Trying z= x^m, z'= mx^m-1, and z''= m(m-1)x^m-2. Putting those into the equation, x²(m(m- 1)x^{m- 2})+ 3x(mx^m-1)- 3x^m= (m(m- 1)- 3m+ 3)x^m= 0. Since x can be non-zero, in order that this not be identically 0 we must have m(m- 1)- 3m+ 3= m²- 4m+ 3= (m- 1)(m- 3)= 0. Yes, m= 1 and m= 3 are roots which means that two independent solutions to the associated homogeneous equation are x and x³. I don't know what you mean by "which doesn't solve the adjoint equal to 0". m= 2 give z= x so that z'= 1 and z''= 0. utting those into the equation, x²(0)+ 3x(1)- 3x= 0 is certainly true. The general solution to the associated homogeneous equation is z= Ax+ Bx³.

Thanks, got mixed up with the definition of the adjoint operator. Sorted now, cheers!