Question

Find a Particular Solution of y'''-4y'=t+3cos(t)+e^-2t

Answer 1

Now the given initil value problem is   y’’’ - 4y’=t + 3cost + e^-2t

So let us take the homogeneous equation for solving

So y’’’ - 4y’= 0    -------------------(1)

Now let us find the characteristic equation for homogeneous equation by assuming the solution y = e^rt not equla to Zero.

So y’ = r e^rt

      y’’ = r²e^rt    

     y’’’ = r³e^rt   

so substitute in (1)

r³e^rt -4re^rt = 0

e^rt (r³ - 4r) = 0

e^rt is not equal to zero so r³ - 4r = 0

• r(r² –4) =0
• r (r–2)(r+2) =0
•   r(r - 2) (r + 2) = 0
• r₁ = 0, r₂ = 2 and r₃=-2
• so when the characteristic equation roots are real and different then the homogeneous solution is in the form as given below.

Yc = C₁e^r1t + C₂ e^r2t + C₃^r3t

Yc = C₁ + C₂e^2t + C₃e^-2t