Solve the given thirdorder differential equation by variatio

Solve the given third-order differential equation by variation of parameters.

y\'\'\'-3y\'\'-y\'+3y=e^5x

Solution

y\' = 3y

y\' - 3y = 0

Multiply throughout by e^-3t,

dy/dx(e^-3t) - 3ye^-3t = 0

The lhs is exactly what is found using the product rule on d/dx(ye^-3t)

So
d/dx(ye^-3t) = 0

Integrate both sides
ye^-3t = C
where C is a constant of integration

Multiply throughout by e^3t
y = Ce^3t

Boundary conditions being y=8 when t=0
8=Ce^0t

So
C=8

Final result
y=8e^3t