How do you find all solutions of the differential equation #(d^2y)/(dx^2)=x^-2#?

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Answer 1 · 2016-12-29T23:38:00

#y=-ln(absx)+C_1x+C_2#

Note that when we integrate things like #dx^2# we will be left with #dx#, so we will need to integrate this twice.

Separating variables:

#d^2y=x^-2dx^2#

Integrating:

#intd^2y=intx^-2dx^2#

This is analogous to doing #intdy=intx^-2dx=>y=-x^-1+C_1#, but a #dx# term will still be left.

#dy=(-x^-1+C_1)dx#

Integrating again:

#intdy=int(-1/x+C_1)dx#

Add another constant:

#y=-ln(absx)+C_1x+C_2#