2nd order diff equation

[gonch76]

I have put down the whole question and my answers which i would appreciate if someone could comment:
Part 1.1
Find the general solution of the equation (d^2 z)/(dx^2 ) - 4 * dz/dx + 13z = 0
Put into form: a * m^2 + b * m + c
∴ m^2 - 4m + 13 = 0 a = 1, b = -4, c = 13
Then:
m_(1,2 = (-b±√(b^2-4ac))/2a)
∴m= (-(-4)±√(〖(-4)〗^2-(4*1*13)))/(2*1)
∴m= (4±√(16-52))/2
∴m= (4±√((-36)))/2
∴m= (4±6j)/2
∴m= (2 (2 ±3j))/2 Note: the 2’s cancel
Then
e^(∝x )* (Acos (βx)+ Bsin (βx) )
∝ =2
β=3
∴ z(x)= e^(2x )* (Acos (3x)+ Bsin (3x) )

Part 1.2
Find the particular solution of the form u(x)= C* e^2x
for the differential equation (d^2 u)/(dx^2 ) - 4 * du/dx + 13u = 〖27e〗^2x
∴u(x)= C* e^2x du/dx=C* 〖2e〗^2x (d^2 u)/(dx^2 )=C* 〖4* e〗^2x Note: 27 is ignored as it is a constant
Then replace:
∴ C*4* e^2x + (- 4) * C * 〖2e〗^2x + 13 * C * e^2x=〖27e〗^2x
∴ C*e^2x * (4 + (-8) + 13)=〖27e〗^2x Note: e^2x cancel out
∴ C*9=27 ∴ C=27/9 = 3
Substituting ‘C’ into the equation gives:
∴ u(x)= 3 * e^2x


Part 1.3
Find the general solution of the differential equation (d^2 u)/(dx^2 ) - 4 * du/dx + 13u = 〖27e〗^2x
General solution when the equation is equal to zero + the particular solution = general solution when equation equals the same value as the particular.

The general solution can be expressed as: y(x)= u(x)+ z(x)
∴y(x)= 3 * e^2x + e^(2x )* (Acos (3x)+ Bsin (3x) )

Note: I believe that the above equation ((d^2 u)/(dx^2 ) - 4 * du/dx + 13u = 〖27e〗^2x) is meant to be dy/dx, not du/dx.


any comments gratefully received.. Cheers.

 

[vela]

Looks good to me. (I didn't check your calculations really closely, so it's possible there's an algebra mistake lurking in there.)

 

[gomunkul51]

correct.
and you should stick to only y(x), z(x), or u(x) in the ODE if it stays the same ;)