Linear Differential Equations and Linear Operator Problem

[whitegirlandrew]

Homework Statement

I'm not sure how to approach this. The question involves linear operators and a non-homogenous differential equation.
Here is the question:
https://s15.postimg.org/cdmw80157/Capture.png

Homework Equations

They are given in the question

The Attempt at a Solution

I really have no idea on how to approach it. I was thinking something along the lines of setting c1*y1+c2*y2+c3*y3 as a linear combination and setting them equal to the non-homongenous term g(x). Then i can find the values of c1,c2,c3.

 

[LCKurtz]

Homework Statement

I'm not sure how to approach this. The question involves linear operators and a non-homogenous differential equation.
Here is the question:
https://s15.postimg.org/cdmw80157/Capture.png

Homework Equations

They are given in the question

The Attempt at a Solution

I really have no idea on how to approach it. I was thinking something along the lines of setting c1*y1+c2*y2+c3*y3 as a linear combination and setting them equal to the non-homongenous term g(x). Then i can find the values of c1,c2,c3.

[Edited] You want to take a linear combination of the $L(y_i)$'s (the right hand side of the equations) equal to $g(x)$, not the $y_i$'s themselves. Then use your constants to build the particular solution from the $y_i$'s.

 

[whitegirlandrew]

I did L(c1y1+c2y2+c3y3)=g(x)=c1*L(y1)+c2*L(y2)+c3*L(y3)
Then I sub in the L(y1), L(y2), and L(y3) for what was given to me. This leads me to find the constants. Then c1=10/12, c2=0, and c3=5/6.
However this was wrong.

 

[LCKurtz]

I did L(c1y1+c2y2+c3y3)=g(x)=c1*L(y1)+c2*L(y2)+c3*L(y3)
Then I sub in the L(y1), L(y2), and L(y3) for what was given to me. This leads me to find the constants. Then c1=10/12, c2=0, and c3=5/6.
However this was wrong.

I corrected my reply while you were looking at it. Sorry. See if it makes sense now.

 

[whitegirlandrew]

Ah I see. I got the question wrong so I don't know the answer. I will try this once the solution comes out. Thanks for the help.