Ordinary Differential Equation = ODE [Principle of superposition] Consider the f

Question

Ordinary Differential Equation = ODE

[Principle of superposition] Consider the following second order, linear, homogeneous ODE: A(x)f"(x) + B{x)f'(x) + C(x)f(x) = 0, where A(x), B(x), C(x) are some functions. Suppose that f 1(x), f 2(x) are solutions of (1). a) Show that for any numbers C 1,C 2 the function f(x) = c 1 f 1(x) + C 2 f 2(x) is also a solution of (1). b) Suppose that f 1 (0)f'2(0) - f 1'(0)f 2 (0) 0. Prove that there exists a unique solution of (1) of the form f(x) = c 1 f 1(x) + c 2 f 2(x) such that f(0) = 7 and f'(0) = 23. Let D(x) be a function and let fo(x) be any solution of the following non-homogeneous ODE A(x)f"(x) + B(x)f(x) + C(x)f(x) = D(x). c) Prove that there exists a unique solution of the ODE (2) of the form f(x) = f o (x) + c 1 f1(x) + C 2 f 2(x) such that f(0) = 7 and f'(0) = 23.

Explanation / Answer

a) Let f(x) = c1f1(x) + c2f2(x) then :

A(x)f''(x)+B(x)f'(x)+C(x)f(x) = c1(A(x)f1''(x)+B(x)f1'(x)+C(x)f1(x))+c2(A(x)f2''(x)+B(x)f2'(x)+C(x)f2(x))=c1*0+c2*0=0

So f(x) is also a solution

b)

c1f1(0)+c2f2(0)=7

c1f1'(0)+c2f2'(0)=23

The system can be written with the matrix

A=

f1(0) f2(0)

f1'(0) f2'(0)

such as A[c1 c2]^T = [7 23]^T

The determinant of A is f1(0)f2'(0)-f1'(0)f2(0), which is !=0 by hypothesis, so the system is consistent and there exist an unique solution of (1) of the form f(x)=c1f1(x)+c2f2(x) such that f(0)=7 and f'(0)=23

It's not clear wether or not we should solve the system, if we need, here it is :

A^(-1)=1/(f1(0)f2'(0)-f1'(0)f2(0))*

f2'(0) -f2(0)

-f1'(0) f1(0)

So [c1 c2]^T = 1/(f1(0)f2'(0)-f1'(0)f2(0))[7f2'(0)-23f2(0) , -7f1'(0)+23f1(0)]

Which gives :

c1 = (7f2'(0)-23f2(0))/(f1(0)f2'(0)-f1'(0)f2(0))

c2 = (-7f1'(0)+23f1(0))/(f1(0)f2'(0)-f1'(0)f2(0))

c)

Suppose f(x) = f0(x) + c1f1(x)+c2f2(x)

Then :

A(x)f'''(x)+B(x)f'(x)+C(x)f(x)

= (A(x)f0''(x)+B(x)f0'(x)+C(x)f0(x)) + c1(A(x)f'1''(x)+B(x)f1'(x)+C(x)f1(x))+c2(A(x)f2''(x)+B(x)f2'(x)+C(x)f2(x))

= D(x) + 0 + 0

So f(x) is solution to (2)

And we need :

f(0)=7=f0(0)+c1f1(0)+c2f2(0)

f'(0)=23=f0'(0)+c2f1'(0)+c2f2'(0)

Which is the sytem A[c1 c2]^T = [7-f0(0),23-f0'(0)]^T

As before, the system is consistent since A is invertible.

If you need to solve the system, then :

c1 = (f0'(0)f2(0)-23f2(0)-f0(0)f2'(0)+7f2'(0))/(f1(0)f2'(0)-f1'(0)f2(0))

c2 = (-f0'(0)f1(0)+23f1(0)+f0(0)f1'(0)-7f1'(0))/(f1(0)f2'(0)-f1'(0)f2(0))

So there exist an unique solution of the ODE (2) such that f(0)=7 and f'(0)=23

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