Verify that y1(t)=t^2 and y2(t)=t^-1 are two solutions of the differential equation t^2y''-2y=0 for t0. Then show that y=C1t^2+C2t^-1 is also a solution of this equation for any C1 and C2.....need help with second part(c1 and c2)

Question

Verify that y1(t)=t^2 and y2(t)=t^-1 are two solutions of the differential equation t^2y''-2y=0 for t>0. Then show that y=C1t^2+C2t^-1 is also a solution of this equation for any C1 and C2.....need help with second part(c1 and c2)

turingtest · Answer

doesn't the principle of superposition cover this?
do you have to work without that?

turingtest · Answer

ok fine, just plugging it in should work too

turingtest · Answer

$$t^2y''-2y=0$$$$t^2(2c_1+2c_2t^{-3})-2(c_1t^2+c_2t^{-1})$$$$=2c_1t^2+2c_2t^{-1}-2c_1t^2-2c_2t^{-1}=0\huge \checkmark$$and that's it :D

anonymous · Answer

with super position

turingtest · Answer

the principle of superposition states that if y1 and y2 are two linearly independent solutions to a differential equation, then so is any linear combination of those solutions$$y=c_1y_1+c_2y_2$$so there is nothing to do

anonymous · Answer

Im  guessing that show implies to plug it in as you did. Same as the first part.Thanx!

turingtest · Answer

welcome!