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Solve the given initial value problem; x^(2)y" + 15xy' + 49y = 0, y(1) = 3, y'(1) = -21

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**Question**

Solve the given initial value problem; x^(2)y" + 15xy' + 49y = 0, y(1) = 3, y'(1) = -21

**Solution**

Solve x^2 ( d^2 y(x))/( dx^2)+49 y(x)+15 x ( dy(x))/( dx) = 0, such that y(1) = 3 and y'(1) = -21: 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: 15 x ( d)/( dx)(x^lambda)+x^2 ( d^2 )/( dx^2)(x^lambda)+49 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+14 lambda x^lambda+49 x^lambda = 0 Factor out x^lambda: (lambda^2+14 lambda+49) x^lambda = 0 Assuming x!=0, the zeros must come from the polynomial: lambda^2+14 lambda+49 = 0 Factor: (lambda+7)^2 = 0 Solve for lambda: lambda = -7 or lambda = -7 The multiplicity of the root lambda = -7 is 2 which gives y_1(x) = c_1/x^7, y_2(x) = (c_2 log(x))/x^7 as solutions, where c_1 and c_2 are arbitrary constants. The general solution is the sum of the above solutions: y(x) = y_1(x)+y_2(x) = c_1/x^7+(c_2 log(x))/x^7 Solve for the unknown constants using the initial conditions: Compute ( dy(x))/( dx): ( dy(x))/( dx) = ( d)/( dx)(c_1/x^7+(c_2 log(x))/x^7) = -(7 c_1)/x^8+c_2/x^8-(7 c_2 log(x))/x^8 Substitute y(1) = 3 into y(x) = c_1/x^7+(c_2 log(x))/x^7: c_1 = 3 Substitute y'(1) = -21 into ( dy(x))/( dx) = -(7 c_1)/x^8+c_2/x^8-(7 c_2 log(x))/x^8: c_2-7 c_1 = -21 Solve the system: c_1 = 3 c_2 = 0 Substitute c_1 = 3 and c_2 = 0 into y(x) = c_1/x^7+(c_2 log(x))/x^7: Answer: | | y(x) = 3/x^7

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