By following the suggested steps, find a particular solution, y_(p) by variation of parameters t^2y''-2y=9t^2+6
t^2r(r-1)t^(r-2))-2t^r=0 r^2-r-2=0 Therefore, y_(c)=C1t^(r1)+C2t^(r2) where r1<=r2. Let y1=t^(r1) and y2=t^(r2).
Step 2: To apply the variation of parameters, find the standard form. y''+p(t)y'+q(t)y=g(t)
Step 3: Apply the variation of parameters to find y_(p)=u1y1+u2y2
[[y1,y2,y2'], [y1',u1'], [u2']]=[[0], [y1,y2,-y2], [-y1',0,y1]] g(t)
u1=int (-y2g(t))/(W(y1,y2))dt u2=int (y1g(t))/(W(y1,y2))dt
Step 4: Calculate y_(p)(0.8) Include at least 3 decimal places in each step in your calculation. Use π=3.14159 and e=2.71828. Type your answer...