solve the differential equation

Question

4x²y"(x)+8xy'(x)+17y(x)=0  x is greater than 0; subject to y(1)=1 and y'(1)=-1/2 

 

a. solve the d.e

b. determine the values of the parameters from the initial condition

 

Answer 1

The differential equation is 4x² * y''(x) + 8x * y'(x) + 17 * y(x) = 0. Since x > 0, we do no have to worry about the analyticity of this point (x = 0) to see if it is an regular or irregular singular point.

Make the ansatz that the solution is y(x) = x^r

y'(x) = r * x^(r-1)

y''(x) = r * (r-1) * x^(r-2)

Plug into the original equation...

4x² * r * r-1 * x^(r-2) + 8x * r * x^(r-1) + 17 x^r = 0

4r*(r-1) * x^r + 8r * x^r + 17 x^r = 0 --> [ 4r*(r-1) + 8r + 17 ] * x^r = 0

Since x > 0 the only way for this equation to be zero is if the quadratic function of r is zero.

4r² - 4r + 8r + 17 = 0

4r² + 4r + 17 = 0

Using the quadratic equation to solve for x yields...

r = [ -4 ± √ (16 - 4*4*17) ] / 8 = [ -4 ± √(16)(1 - 17) ] / 8 = [ -4 ± √(16)(-16) ] / 8 = [ -4 ± 16i ] / 8 = (-1 ± 4i ) / 2

Since x is imaginary that will state our solutions will need to be recast...

r₁ = (-1 + 4i ) / 2

r₂ = (-1 - 4i ) / 2

y(x) = y(r₁,x) + y(r₂,x) = Ax^{(-1 + 4i) / 2} + Bx^{(-1 - 4i) / 2}

y'(x) = (-1 + 4i) / 2 * A * x^{(-3 + 4i) / 2} + (-1 - 4i) / 2 * B * x^{(-3 - 4i) / 2}

So we need to know how to convert the last two items into something useful so we can solve for the unknown coefficients.

x²ⁱ = e^{(2i * ln(x))}

So we can see that these are representative of exponential solutions with an imaginary argument which can be written as sine and cosine...

x²ⁱ = cos(2ln(x)) + isin(2ln(x))

x^-2i = cos(2ln(x) - isin(2ln(x))

Therefore we can write our solution as ...

y(x) = Ax^{(-1 + 4i) / 2} + Bx^{(-1 - 4i) / 2} = A / √x * [ cos(2ln(x)) + isin(2ln(x)) ] + B / √x * [ cos(2ln(x) - isin(2ln(x)) ]

y(x) = Ax^{(-1 + 4i) / 2} + Bx^{(-1 - 4i) / 2} = 1 / √x * [ (A+B) * cos(2ln(x)) + (A-B) * isin(2ln(x))

y'(x) = (-1 + 4i) / 2 * A * x^{(-3 + 4i) / 2} + (-1 - 4i) / 2 * B * x^{(-3 - 4i) / 2} = 1 / √(x³) * [ (A+B) * cos(2ln(x)) + (A-B) * isin(2ln(x))

Since x > 0, we now solve for A and B

y(1) = 1 = 1 / √1 * [ (A+B) * cos(2ln(1)) + (A-B) * isin(2ln(1)) ] = (A+B) * cos(0) + (A-B) * isin(0) = A+B

y'(1) = -1/2 = 1 / √(1)³ * [ ((-1 + 4i) / 2 * A + (-1 - 4i) / 2 * B) * cos(2ln(1)) + ((-1 + 4i) / 2 * A - (-1 - 4i) / 2 * B) * isin(2ln(1)) ] = -(A+B) + 2i(A-B)

y(1) = 1 = A+B

y'(1) = -1/2 = -(A+B) + 2i(A-B)

Solving this system of equation gives...

1/2 = 2i(A-B) --> -i/4 = A-B, A+B = 1

2A = 1 - i/4 --> A = 1/2 - i/8

-2B = -i/4 - 1 --> B = -(i/8 + 1/2)

So...

y(x) = 1 / √x * [ (A+B) * cos(2ln(x)) + (A-B) * isin(2ln(x)) ] = 1 / √x * [ cos(2ln(x)) + 1/4 * sin(2ln(x)) ] for x > 0