use variation parameters to find a general solution to the differential equation given that the function y1 and y2 are linearly independent solution
ty''-(t+1)y'+y=t^2; y1=e^t,y2=t+1
1) divide differential equation by t [tex]\Large {y''- \frac{(t+1)}{t} y'+\frac1ty=t}[/tex]
2)Solve this system of equations for [tex]v_1'[/tex] and [tex]v_2'[/tex] [tex]v_1'\left(e^t\right) +v_2'\left(t+1\right)=0 \\ v_1'\left(e^t\right) +v_2'\left(1\right)\ \ \ \ \ =t[/tex]
you get [tex]{v_1}'=\left(t+1\right)/e^t, \ \ {v_2}'=-1[/tex]
