Help finding general solution for 2nd order linear DE

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SUMMARY

The general solution for the second-order linear differential equation \( t^2y'' - 2y = 0 \) is \( y(t) = C_1t^2 + C_2t^{-1} \), where \( C_1 \) and \( C_2 \) are constants. The solutions \( y_1(t) = t^2 \) and \( y_2(t) = t^{-1} \) are fundamental solutions derived from the characteristic equation. The independent variable is \( t \), not \( x \), which is a common point of confusion in the discussion.

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Jende
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Homework Statement


Find the general solution for the DE: t2y''-2y=0

Homework Equations


These were given for other parts of the problem so I'm not sure if they're relevant.
y1(t)=t2, y2(t)=t-1, y(1)=-2, y'(1)=-7

The Attempt at a Solution


The t2 at the front was really stumping me and I'm not sure where to begin.

Thanks in advance for any help.
 
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Jende said:

Homework Statement


Find the general solution for the DE: t2y''-2y=0

Homework Equations


These were given for other parts of the problem so I'm not sure if they're relevant.
y1(t)=t2, y2(t)=t-1, y(1)=-2, y'(1)=-7

The Attempt at a Solution


The t2 at the front was really stumping me and I'm not sure where to begin.

Thanks in advance for any help.

The earlier parts are relevant. What are ##y_1(t)## and ##y_2(t)##? That is, what do they have to do with this problem? Once you answer that, what do you know about solutions of linear equations?
 
LCKurtz said:
The earlier parts are relevant. What are ##y_1(t)## and ##y_2(t)##? That is, what do they have to do with this problem? Once you answer that, what do you know about solutions of linear equations?

So it should come out to be: y(x)=C1t2+C2t-1
 
Jende said:
So it should come out to be: y(x)=C1t2+C2t-1

So what should come out that? And does it?
 
Jende said:
So it should come out to be: y(x)=C1t2+C2t-1
The independent variable is t, not x, so the above should be y(t)=C1t2+C2t-1
 

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