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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.