Marks: --/1 Select all values of m for which y = x m is a solution of x 2 y \'\'

Question

Marks: --/1 Select all values of m for which y = xm is a solution of
x2y'' - 5xy' + 8y = 0.
Choose at least one answer. a. -2
b. 1
c. 5
d. 3
e. -4
f. 2
g. 4 Marks: --/1 Select all values of m for which y = xm is a solution of
x2y'' - 5xy' + 8y = 0.
Choose at least one answer. a. -2
b. 1
c. 5
d. 3
e. -4
f. 2
g. 4 Select all values of m for which y = xm is a solution of
x2y'' - 5xy' + 8y = 0.
Choose at least one answer. a. -2
b. 1
c. 5
d. 3
e. -4
f. 2
g. 4 Choose at least one answer. a. -2
b. 1
c. 5
d. 3
e. -4
f. 2
g. 4

Explanation / Answer

Solve ( d^2 y(x))/( dx^2) x^2-5 ( dy(x))/( dx) x+8 y(x) = 0: Assume a solution to this Euler-Cauchy equation will be proportional to x^lambda for some constant lambda. Substitute y(x) = x^lambda into the differential equation: x^2 ( d^2 )/( dx^2)(x^lambda)-5 x ( d)/( dx)(x^lambda)+8 x^lambda = 0 Substitute ( d^2 )/( dx^2)(x^lambda) = (lambda-1) lambda x^(lambda-2) and ( d)/( dx)(x^lambda) = lambda x^(lambda-1): lambda^2 x^lambda-6 lambda x^lambda+8 x^lambda = 0 Factor out x^lambda: (lambda^2-6 lambda+8) x^lambda = 0 Assuming x!=0, the zeros must come from the polynomial: lambda^2-6 lambda+8 = 0 Factor: (lambda-4) (lambda-2) = 0 Solve for lambda: lambda = 2 or lambda = 4 The root lambda = 2 gives y_1(x) = c_1 x^2 as a solution, where c_1 is an arbitrary constant. The root lambda = 4 gives y_2(x) = c_2 x^4 as a solution, where c_2 is an arbitrary constant. The general solution is the sum of the above solutions: Answer: | | y(x) = y_1(x)+y_2(x) = c_1 x^2+c_2 x^4hencem=2,1

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