A calculus problem by Rocco Dalto

Using the substitution $z = \ln(x - \ln(x+1))$ yields the differential equation $\frac{d^2y}{dz^2} - \frac{dy}{dz} + y \; = \; 0$ and we note that $z=0$ when $x=a$ . Note that there is only one positive value $a$ for which $\ln(a+1) = a-1$ , and so $a$ is well-defined. This equation has the solutions $e^{mz}$ where $m = \tfrac12 \pm \tfrac12i\sqrt{3}$ . Thus we obtain linearly independent solutions $\begin{array}{rcl} y_1(x) & = & e^{\frac12z} \cos(\tfrac12\sqrt{3}z) \; = \; \sqrt{x - \ln(x+1)}\cos\big(\tfrac12\sqrt{3} \ln(x - \ln(x+1))\big) \\ y_2(x) & = & e^{\frac12z} \sin(\tfrac12\sqrt{3}z) \; = \; \sqrt{x - \ln(x+1)}\sin\big(\tfrac12\sqrt{3} \ln(x-\ln(x+1))\big) \end{array}$ Since $z'(a) \,=\, \tfrac{a}{a+1}$ , it is easy to check that $y_1,y_2$ as given above are the required solutions of the differential equation. But then $y_1(x)^2 + y_2(x)^2 = x - \ln(x+1)$ , and so $\int_0^{e-1} \big( y_1(x)^2 + y_2(x)^2\big)\,dx \; = \; \int_0^{e-1} (x - \ln(x+1))\,dx \; = \; \tfrac12(e^2 - 2e -1) \; = \; \boxed{0.476246}$

Prior to solving the differential $\dfrac{d^2y}{dx^2} - \dfrac{1}{x(x + 1)} \dfrac{dy}{dx} + \dfrac{x^2}{(x + 1)^2 (x - \ln(x + 1))^2} y = 0$ ,
I am going to demonstrate how I constructed the differential, and reduced it to a second order differential with constant coefficients. If you reviewed(or proved) the statement(in the Hint ), as well as my second statement which outlines the construction, you will be familiar with what follows. $$

Let $P(x) = -\dfrac{1}{x (x + 1)} = -(\dfrac{1}{x} - \dfrac{1}{x + 1})$ and let the constant $j = -2 \implies$ $$

$\dfrac{dQ}{dx} - \dfrac{2}{x (x + 1)} Q = -2 Q^{\dfrac{3}{2}} \implies$ $$

$Q^{-\dfrac{3}{2}} \dfrac{dQ}{dx} - \dfrac{2}{x (x + 1)} Q^{-\dfrac{1}{2}} = -2.$ $$

Let $W = Q^{-\dfrac{3}{2}} \implies -2 \dfrac{dW}{dx} = Q^{-\dfrac{3}{2}} \dfrac{dQ}{dx} \implies$ $$

$\dfrac{dW}{dx} + \dfrac{1}{x (x + 1)} W = 1 \implies$ $$

$\dfrac{d}{dx}(\dfrac{x}{x + 1} * W) = \dfrac{x}{x + 1} \implies$ $$

$\dfrac{x}{x + 1} * W = \int (1 - \dfrac{1}{x + 1}) \: dx + C = x - \ln(x + 1) + C \implies$ $$

$W = (\dfrac{x + 1}{x}) (x - \ln(x + 1)) + (\dfrac{x + 1}{x}) * C.$ $$

Let $C = 0 \implies W = (\dfrac{x + 1}{x}) (x - \ln(x + 1))$ $$

$W = Q^{-\dfrac{1}{2}} \implies Q = \dfrac{x^2}{(x + 1)^2 (x - \ln(x + 1))^2}$ $\implies$ $$

$z(x) = \int \sqrt{Q(x)} \: dx = \int (\dfrac{1 - 1/(x + 1)}{x - \ln(x + 1)^2}) \: dx = \ln(x - ln(x + 1))$

Now we have constructed our second order differential $\dfrac{d^2y}{dx^2} - \dfrac{1}{x(x + 1)} \dfrac{dy}{dx} + \dfrac{x^2}{(x + 1)^2 (x - \ln(x + 1))^2} y = 0$ and the substitution $z = \ln(x - ln(x + 1))$ will reduce this differential to a second order differential with constant coefficients, since $\dfrac{Q' + 2PQ}{Q^{3}{2}} = -2.$ $$

Now we begin solving the differential $\dfrac{d^2y}{dx^2} - \dfrac{1}{x(x + 1)} \dfrac{dy}{dx} + \dfrac{x^2}{(x + 1)^2 (x - \ln(x + 1))^2} y = 0.$ $$

From above, let $z = \ln(x - ln(x + 1)) \implies \dfrac{dy}{dx} = (\dfrac{x}{x + 1}) (\dfrac{1}{x - \ln(x + 1)}) \dfrac{dy}{dz}$ $$

and, $$

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

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

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

$\dfrac{d^2y}{dz^2} - \dfrac{dy}{dx} + y = 0 \implies m^2 - m + 1 = 0 \implies m = \dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2} \implies$ $$

$y(z) = e^\dfrac{z}{2} * (c1 * cos(\dfrac{\sqrt{3}}{2} z) + c2 * sin(\dfrac{\sqrt{3}}{2} z))$ $$

$z = \ln(x - ln(x + 1)) \implies$ $$

The general solution is: $y(x) = \sqrt{x - \ln(x + 1)} * (c1 * cos(\dfrac{\sqrt{3}}{2} * (\ln(x - \ln(x + 1))) + c2 * sin(\dfrac{\sqrt{3}}{2} * (\ln(x - \ln(x + 1))))$ $$

$\therefore$ The two linear independent solutions are: $$

$y1(x) = c1 * \sqrt{x - \ln(x + 1)} * cos(\dfrac{\sqrt{3}}{2} * (\ln(x - \ln(x + 1)))$ $$

$y2(x) = c2 * \sqrt{x - \ln(x + 1)} * sin(\dfrac{\sqrt{3}}{2} * (\ln(x - \ln(x + 1)))$ $$

Using the initial conditions $y_{1}(a) = 1, \: y_{1}'(a) = \dfrac{a}{2 * (a + 1)},$ $y_{2}(a) = 0, \: y_{2}'(a) = \dfrac{\sqrt{3}}{2} * (\dfrac{a}{a + 1}) \implies c1 = c2 = 1.$

$y1(x)$ and $y2(x)$ are clearly linear independent since $\dfrac{y2}{y1} = tan(\dfrac{\sqrt{3}}{2} * (\ln(x - \ln(x + 1))) \not=$ constant. $$

and,

$\int_{0}^{e - 1} y1(x)^2 + y2(x)^2 \: dx = \int_{0}^{e - 1} (x - \ln(x + 1)) \: dx$ $$

For $\int \ln(x + 1) \: dx$ $$

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

$\int \ln(x + 1) \: dx = x \ln(x + 1) - \int (1 - \dfrac{1}{x + 1}) \:dx = x * \ln(x + 1) - x + \ln(x + 1) =$ $$ $(x + 1) * \ln(x +1) - x$ $$

$\implies$ $$

$\int_{0}^{e - 1} (x - \ln(x + 1)) \: dx = \dfrac{1}{2} (x) (x + 2) - (x + 1) * \ln(x + 1)|_{0}^{e - 1} = \dfrac{e^2 - 2 e - 1}{2} = .476246$ to six decimal places.