A calculus problem by Rocco Dalto

To avoid using a power series solution we reduce $x^2 \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + 2y = 0$ to a second order differential with constant coefficients by letting $$

$z = \ln(x) \implies \dfrac{dy}{dx} = \dfrac{dy}{dz} * \dfrac{1}{x}$ $$

and,

$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(\dfrac{dy}{dz} * \dfrac{1}{x}) = \dfrac{dy}{dz}(-\dfrac{1}{x^2}) + \dfrac{1}{x} \dfrac{d}{dx}(\dfrac{dy}{dz}) = \dfrac{-1}{x^2} \dfrac{dy}{dz} + \dfrac{1}{x} \dfrac{d}{dz}(\dfrac{dy}{dz}) \dfrac{1}{x} =$

$\dfrac{1}{x^2} \dfrac{d^2y}{dz^2} - \dfrac{1}{x^2} \dfrac{dy}{dz}$ $$

Replacing $\dfrac{dy}{dx}$ and $\dfrac{d^2 y}{dx^2}$ in the differential above we obtain: $$

$\dfrac{d^2 y}{dz^2} - 2 \dfrac{dy}{dz} + 2y = 0$ $$

Assuming your familiar with solving second order differential equations with constant coefficients we obtain: $$

$m^2 - 2m + 2 = 0 \implies m1,m2 = 1 \pm i \implies y(z) = e^z * (c1 * cos(z) + c2 * sin(z))$ $$

$z = \ln(x) \implies y(x) = x * (c1 * cos(\ln(x)) + c2 * sin(\ln(x)))$

and, $$

$\dfrac{dy}{dx} = (c2 + c1) * cos(\ln(x)) + (c2 - c1) * sin(\ln(x))$ $$

Using the initial conditions $y(1) = \dfrac{3}{2}, y'(1) = 2$ we obtain $c1 = c2 = 1 \implies$

$y(x) = x * (\dfrac{3}{2} * cos(\ln(x)) + \dfrac{1}{2} * sin(\ln(x)))$ $$

The area $A = \int_{1}^{e^{2\pi}} x * (\dfrac{3}{2} cos(\ln(x)) + \dfrac{1}{2} sin(\ln(x))) \: dx$ $$

For $\int sin(\ln(x)) \: dx$ $$

Let $u = sin(\ln(x)), dv = dx \implies du = \dfrac{1}{x} cos(\ln(x)), \: and \: v = x$ $\implies$ $$

$\int sin(\ln(x)) \: dx = x * sin(\ln(x)) - \int cos(\ln(x)) dx$ $$

Let $u = cos(\ln(x)), dv = dx \implies du = -\dfrac{1}{x} sin(\ln(x)), \: and \: v = x$ $\implies$ $$

$\int sin(\ln(x)) \: dx = x * sin(\ln(x)) - x * cos(\ln(x)) - \int sin(\ln(x)) \: dx \implies$ $$

$2 * \int sin(\ln(x)) \: dx = x * (sin(\ln(x)) - cos(\ln(x))) \implies$ $$

$\int sin(\ln(x)) \: dx = \dfrac{x}{2} * (sin(\ln(x)) - cos(\ln(x)))$ $$

For $\int cos(\ln(x)) \: dx$ $$

Let $u = cos(\ln(x)), dv = dx \implies du = -\dfrac{1}{x} sin(\ln(x)), \: and \: v = x$ $\implies$ $$

$\int cos(\ln(x)) \: dx = x * cos(\ln(x)) + \int sin(\ln(x)) \: dx = \dfrac{x}{2} * (sin(\ln(x)) + cos(\ln(x)))$ $$ $\implies$

$\int sin(\ln(x)) + cos(\ln(x)) \: dx = x * sin(\ln(x))$ $$

For $\int x * (sin(\ln(x)) + cos(\ln(x))) \: dx$ $$

Let $u = x, dv = sin(\ln(x)) + cos(\ln(x)) \: dx \implies du = dx \: and \: v = x * sin(ln(x)) \implies$ $$

$\int x * (sin(\ln(x)) + cos(\ln(x))) \: dx = x^2 * sin(\ln(x)) - \int x * sin(\ln(x)) \: dx$ $$

Let $u = x, dv = sin(\ln(x)) \: dx \implies du = dx \: and \: v = \dfrac{x}{2} * (sin(ln(x)) - cos(\ln(x))) \implies$ $$

$\int x * sin(\ln(x)) \: dx + \int x * cos(\ln(x)) \: dx =$ $$

$x^2 * sin(\ln(x)) - (\dfrac{x^2}{2} * (sin(\ln(x)) - cos(\ln(x))) + \dfrac{1}{2} * \int x * sin(\ln(x)) \: dx - \dfrac{1}{2} * \int x * cos(\ln(x)) \: dx$ $$

$\implies$

$\int \dfrac{3}{2} * x * cos(\ln(x)) + \dfrac{1}{2} * x * sin(\ln(x)) \: dx =$ $$

$\dfrac{x^2}{2} * (sin(\ln(x)) + cos(\ln(x)))$ $$

$\therefore$ the area $A = \dfrac{x^2}{2} * (sin(\ln(x)) + cos(\ln(x)))|_{1}^{e^{2\pi}} =$ $$

$\dfrac{1}{2} * (e^{4\pi} - 1) = \dfrac{a}{b} * (e^{c\pi} - a) \implies$ $$

$a + b + c = 7.$

Trying a solution of the form $y=x^m$ works if $m(m-1)-m+2=0$ So that $m=1\pm i$ . The initial conditions give the solution $y=\tfrac14(3-i)x^{1+i}+\tfrac14(3-i )x^{1-i}$ So the desired integral is $\left[ \frac{3-i}{4(2+i)}x^{2+i}+\frac{3+i}{4(2-i )}x^{2-i}\right]_1^{e^{2\pi}} =\tfrac12(e^{4\pi}-1)$ Making the answer $\boxed{7}$ .

You know you can avoid cumbersome steps by reducing the area expression by using given data itself: (3d/dx(xy) − d/dx(x^2(dy/dx)))/5)= \int_mathrm{y}\,\mathrm{d}x