Answer to Question #97642 in Differential Equations for Niharika

Question

d^2y/d^2 = 1/x(x+1) + cosec^2x

Answer 1

We need to solve the equation

$$y''(x) = \frac{1}{{x(x + 1)}} + \frac{1}{{{{\sin }^2}x}}$$

(we used that ${\csc ^2}x = \frac{1}{{{{\sin }^2}x}}$)

To solve it, we need to integrate it twice. Let's do the first integration

$$y'(x) = {C_1} + \int {\frac{{dx}}{{x(x + 1)}}} + \int {\frac{{dx}}{{{{\sin }^2}x}}}$$

To find the integral $\int {\frac{{dx}}{{x(x + 1)}}}$ let's represent the fraction $\frac{1}{{x(x + 1)}}$ as a sum of simple fractions $\frac{1}{{x(x + 1)}} = \frac{1}{x} - \frac{1}{{x + 1}}$ then

$$\int {\frac{{dx}}{{x(x + 1)}}} = \int {\frac{{dx}}{x}} - \int {\frac{{dx}}{{x + 1}}} = \ln \left| x \right| - \ln \left| {x + 1} \right|$$

The nex integral $\int {\frac{{dx}}{{{{\sin }^2}x}}}$ can be found if we you notice that

$$\frac{{d\cot (x)}}{{dx}} = \frac{{ - \sin x}}{{\sin x}} - \frac{{\cos x}}{{{{\sin }^2}x}}\cos x = - 1 - \frac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}} = - \frac{1}{{{{\sin }^2}x}}$$

It means that

$$\int {\frac{{dx}}{{{{\sin }^2}x}}} = - \cot x$$

We get

$$y'(x) = {C_1} + \ln \left| x \right| - \ln \left| {x + 1} \right| - \cot x$$

Do the integration again

$$y(x) = {C_1}x + {C_2} + \int {\ln \left| x \right|dx} - \int {\ln \left| {x + 1} \right|dx} - \int {\cot xdx}$$

First integral can be calculated by parts. Let $x > 0$ we get

$$\int {\ln xdx} = x\ln x - \int {x \cdot \frac{{dx}}{x}} = x(\ln x - 1)+c$$

For $x < 0$ using $\xi = - x$ we get

$$\int {\ln ( - x)dx} = - \int {\ln \xi d\xi } = - \xi (\ln \xi - 1)+c = x(\ln ( - x) - 1)+c$$

Thus

$$\int {\ln \left| x \right|dx} = x(\ln \left| x \right| - 1)+c$$

The second integral can be reduced to the first using $\eta = x + 1$

$$\int {\ln \left| {x + 1} \right|dx} = \int {\ln \left| \eta \right|d\eta }$$

Thus

$$\int {\ln \left| {x + 1} \right|dx} = (x + 1)(\ln \left| {x + 1} \right| - 1)+c$$

The last integral

$$\int {\cot xdx} = \int {\frac{{\cos x}}{{\sin x}}dx} = \int {\frac{{d(\sin x)}}{{\sin x}}} = \ln \left| {\sin x} \right|+c$$

And we get (we omit some terms because they can be included in ${C_1}x$ and ${C_2}$)

$$y(x) = {C_1}x + {C_2} + x\ln \left| x \right| - (x + 1)\ln \left| {x + 1} \right| - \ln \left| {\sin x} \right|$$