Question

Find a general solution to the Cauchy-Euler equation

\(\begin{array}{l}{x^3}{y^{\prime \prime \prime }} - 2{x^2}{y^{\prime \prime }} - 5x{y^\prime } + 5y = {x^{ - 2}},\\x > 0,\end{array}\)

given that \(\left\{ {x,{x^5},{x^{ - 1}}} \right\}\) is a fundamental solution set to the corresponding homogeneous equation.

Short answer

The general solution for the given equation:

\(y(x) = {c_1}x + {c_2}{x^5} + {c_3}{x^{ - 1}} - \frac{1}{{21}}{x^{ - 2}}\)

Step-by-step solution

Determining the solution set to the corresponding homogeneous equation.

The first step is to divide the given equation by \({x^3}\) to obtain the standard form.

\({y^{\prime \prime \prime }} - \frac{2}{x}{y^{\prime \prime }} - \frac{5}{{{x^2}}}{y^\prime } + - \frac{5}{{{x^3}}}y = {x^{ - 5}}\)

Now we see that\(g(x) = {x^{ - 5}}\). Since \(\left\{ {x,{x^5},{x^{ - 1}}} \right\}\) is a fundamental solution set the solution to the corresponding homogeneous equation is:

\({y_h}(x) = {c_1}x + {c_2}{x^5} + {c_3}{x^{ - 1}}\)

And we can obtain a particular solution of the form.

\({y_p}(x) = {V_1}(x)x + {V_2}(x){x^5} + {V_3}(x){x^{ - 1}}.\)

Determining the function \({V_1},{V_2}\) and\({V_3}\)

Let’s determine the functions \({V_1},{V_2}\)and\({V_3}\). First, we must evaluate the four determinants:

\(\begin{array}{c}W\left[ {x,{x^5},{x^{ - 1}}} \right](x) = \left| {\begin{array}{*{20}{c}}x&{{x^5}}&{{x^{ - 1}}}\\1&{5{x^4}}&{ - {x^{ - 1}}}\\0&{20{x^3}}&{2{x^{ - 3}}}\end{array}} \right| = x\left| {\begin{array}{*{20}{c}}{5{x^4}}&{ - {x^{ - 2}}}\\{20{x^3}}&{2{x^{ - 3}}}\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}{{x^5}}&{{x^{ - 1}}}\\{20{x^3}}&{2{x^{ - 3}}}\end{array}} \right|\\ = x(10x + 20x) - 1\left( {2{x^2} - 20{x^2}} \right)\\ = 30{x^2} + 18{x^2}\\ = 48{x^2}\end{array}\)

\(\begin{array}{c}{W_1} = {( - 1)^{3 - 1}}\left| {\begin{array}{*{20}{c}}{{x^5}}&{{x^{ - 1}}}\\{5{x^4}}&{ - {x^{ - 2}}}\end{array}} \right| = - {x^3} - 5{x^3} = - 6{x^3}\\{W_2} = {( - 1)^{3 - 2}}\left| {\begin{array}{*{20}{c}}x&{{x^{ - 1}}}\\1&{ - {x^{ - 2}}}\end{array}} \right| = - \left( { - {x^{ - 1}} - {x^{ - 1}}} \right) = 2{x^{ - 1}}\\{W_3} = {( - 1)^{3 - 3}}\left| {\begin{array}{*{20}{c}}x&{{x^5}}\\1&{5{x^4}}\end{array}} \right| = 5{x^5} - {x^5} = 4{x^5}\end{array}\)

Now we can calculate the undetermined functions \({V_1},{V_2}\)and\({V_3}\).

\(\begin{array}{c}{V_1}(x) = \int {\frac{{g(x) \cdot {W_1}}}{{W(x)}}} dx = \int {\frac{{{x^{ - 5}} \cdot \left( { - 6{x^3}} \right)}}{{48{x^2}}}} dx = - \frac{1}{8}\int {{x^{ - 4}}} dx = \frac{1}{{24}}{x^{ - 3}}\\{V_2}(x) = \int {\frac{{g(x) \cdot {W_2}}}{{W(x)}}} dx = \int {\frac{{{x^{ - 5}} \cdot 2{x^{ - 1}}}}{{48{x^2}}}} dx = \frac{1}{{24}}\int {{x^{ - 8}}} dx = - \frac{1}{{168}}{x^{ - 7}}\\{V_3}(x) = \int {\frac{{g(x) \cdot {W_3}}}{{W(x)}}} dx = \int {\frac{{{x^{ - 5}} \cdot 4{x^5}}}{{48{x^2}}}} dx = \frac{1}{{12}}\int {{x^{ - 2}}} dx = - \frac{1}{{12}}{x^{ - 1}}\end{array}\)

Determining the general solution for the given equation.

Substituting this into \({y_p}(x) = {V_1}(x)x + {V_2}(x){x^5} + {V_3}(x){x^{ - 1}}\) gives:

\(\begin{array}{c}{y_p}(x) = \frac{1}{{24}}{x^{ - 3}} \cdot x - \frac{1}{{168}}{x^{ - 7}} \cdot {x^5} - \frac{1}{{12}}{x^{ - 1}} \cdot {x^{ - 1}}\\ = \frac{1}{{24}}{x^{ - 2}} - \frac{1}{{168}}{x^{ - 2}} - \frac{1}{{12}}{x^{ - 2}} = - \frac{1}{{21}}{x^{ - 2}}\end{array}\)

Finally, a general solution to the given equation is:

\(\begin{array}{c}y(x) = {y_h}(x) + {y_p}(x)\\y(x) = {c_1}x + {c_2}{x^5} + {c_3}{x^{ - 1}} - \frac{1}{{21}}{x^{ - 2}}\end{array}\)