Question

In Problems \(1 - 16\)the indicated function \({y_1}(x)\) is a solution of the given differential equation. Use reduction of order or formula\((5)\),as instructed, to find a second solution\({y_2}(x)\).

\(\left( {1 - 2x - {x^2}} \right)y'' + 2(1 + x)y' - 2y = 0;\quad {y_1} = x + 1\).

Short answer

The second solution \({y_2}(x)\)of the differential equation\(\left( {1 - 2x - {x^2}} \right)y'' + 2(1 + x)y' - 2y = 0\) is\( - {x^2} - x - 2\).

Step-by-step solution

Define reduction of order.

To solve a second order differential equation, the reduction of order approach is based on the principle of solving first order differential equations one after the other that have been generated from the original second order equation to simplify the problem.

The general formula is,

\({y_2}(x) = {y_1}(x)\smallint \frac{{{e^{ - \smallint p(x)dx}}}}{{{{\left( {{y_2}(x)} \right)}^2}}}dx.\)

Find the value of\({y_2}(x)\).

We are given that,

\(\left( {1 - 2x - {x^2}} \right)y'' + 2(1 + x)y' - 2y = 0;\quad {y_1} = x + 1\)

This is a second order differential equation. It can be written as,

\(y'' + \frac{{2(1 + x)}}{{1 - 2x - {x^2}}}y' - \frac{2}{{1 - 2x - {x^2}}}y = 0\)

as the form,

\(y'' + p(x)y' + q(x)y = 0\).

with a first solution \({y_1} = x + 1\).

We have to obtain the second solution using the formula for reduction of order as,

\({y_2}(x) = {y_1}(x)\smallint \frac{{{e^{ - \smallint p(x)dx}}}}{{{{\left( {{y_2}(x)} \right)}^2}}}dx\)

\( - \smallint \frac{{2 + 2x}}{{1 - 2x - {x^2}}}dx = - \smallint \frac{{2 + 2x}}{{ - (2 + 2x)(u)}}du = \smallint \frac{1}{{(u)}}du = \ln |u| = \ln \left| {1 - 2x - {x^2}} \right|\)

\({y_2} = (x + 1)\smallint \frac{{{e^{\ln \left| {1 - 2x - {x^2}} \right|}}}}{{{{(x + 1)}^2}}}dx = (x + 1)\smallint \frac{{1 - 2x - {x^2}}}{{{{(x + 1)}^2}}}dx = (x + 1)\smallint \frac{{1 - 2x - {x^2}}}{{{x^2} + 2x + 1}}dx\)

Hence the value of \({y_2}(x)\)is \((x + 1)\smallint \frac{{1 - 2x - {x^2}}}{{{x^2} + 2x + 1}}dx\).

Find the second solution of the given differential equation.

To obtain the integration, we have to use integration by substituting.

Now we can obtain the second solution of the given differential equation as,

\(\begin{aligned}{*{20}{c}}{{y_2} = (x + 1)\left[ {\smallint \frac{2}{{{x^2} + 2x + 1}}dx - \smallint 1dx} \right]}\\{ = (x + 1)\left[ {\smallint \frac{2}{{{{(x + 1)}^2}}}dx - \smallint 1dx} \right]}\\{ = (x + 1)\left[ {\smallint \frac{2}{{{{(x + 1)}^2}}}dx - \smallint 1dx} \right]}\\{ = (x + 1)\left[ {\smallint 2{{(x + 1)}^{ - 2}}dx - \smallint 1dx} \right]}\\{ = (x + 1)\left[ { - 2{{(x + 1)}^{ - 1}} - x} \right]}\\{ = (x + 1)\left[ { - \frac{2}{{x + 1}} - x} \right]}\\{ = - {x^2} - x - 2}\end{aligned}\)

Hence the second solution of the given differential equation is \( - {x^2} - x - 2\).