Find two independent solutions for t0 of the associated homo

Solution

t²x\'\'-2tx+2x=0 { y= e^rx & y\' = r e^rx & y\"= r² e^rx}

So,

t r² e^rx - 2t e^rx + 2 e^rx = 0

e^rx ( t r² - 2t +2 )=0

but ( t r² - 2t  +2 ) cant be \'0\' so e^rx   is equals to 0

( t r² - 2t  +2 ) = 0

t ( r ² -2) +2 =0

t ( r ² - 2) = -2

( r ² -2) = -2 / t

for t > 0 put any value of t which is greater than 0 so put 1 and 2

for r_{1 ,} put t= 1

( r ² -2)= - 2/1

r ² = -2 + 2

r₁=0

for r_{2 ,} putt= 2

( r ² -2)= - 2/2

( r ² -2)= - 1

r ² = -1 + 2

r² = 1

r₂ = 1

For independent homogenous equation

= C₁e^r₁x + C₂e^r₂x

= C₁e^{0 x}+ C₂e^1x

= C₁e^{0 x}+ C₂e^1x

= C_{1 +} C₂e^x  ans...