Differntial Eq Given that y1t t is a known solution of the

Differntial Eq

Given that y_1(t) = t is a known solution of the second order linear differential equation t^2y\" - 4ty\' + 4y = 0, t > 0. Find the general solution of the equation.

Solution

Assume a solution:

y2=y1*v=tv

y2\'=v+tv\'

y2\'\'=2v\'+tv\'\'

Substituting gives:

t^2y2\'\'-4ty2\'+4y2=0

t^2 (2v\'+tv\'\')-4t(v+tv\')+4tv=0

2t^2 v\'+t^3 v\'\'-4t^2 v\'=0

t^3v\'\'-2t^2v\'=0

tv\'\'-2v\'=0

d(v\')/v\'=2dt/t

Integrating gives:

ln(v\')=ln(t^2)+C

v\'=At^2

v=At^3/3+B

Hence general solution is:

y(t)=y1v=t(At^3/3+B)=At^4/3+Bt