Question

(a) Use the general solution given in Example 5 to solve the IVP \(4x'' + {e^{ - 0.1t}}x = 0,x(0) = 1,x'(0) = - \frac{1}{2}\). Also use \(J_0^'(x) = - {J_1}(x)\) and \(Y_0^'(x) = - {Y_1}(x)\) along with Table 6.4.1 or a CAS to evaluate coefficients.

(b) Use a CAS to graph the solution obtained in part (a) for \(0 \le t < \infty \).

Short answer

(a) The general solution is \(x(t) = - 4.78601{J_0}\left( {10{e^{ - t/20}}} \right) - 3.18028{Y_0}\left( {10{e^{ - t/20}}} \right)\).

(b) The graph has been plotted.

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be\({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation hastwo linearly independent solutionsfor a fixed value of\(n\).A Bessel equation of the first kind,indicated by\({J_n}(x)\), is one of these solutions that may be derived usingFrobinous approach.

\(\begin{array}{l}{y_1} = {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} = {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{array}\)

At\(x = 0\), this solution is regular. The second solution, which is singular at\(x = 0\), is represented by\({Y_n}(x)\)and is calleda Bessel function of the second kind.

\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)

Determine the form of the general solution.

Let the given be\(4x'' + {e^{ - 0.1t}}x = 0,x(0) = 1,x'(0) = \frac{{ - 1}}{2}\). That has the general solution,\(x(t) = {c_1}{J_0}\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right) + {c_2}{Y_0}\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\), where\({c_1}\)and\({c_2}\)are arbitrary constants The given DE has the form of\(mx'' + k{e^{ - \alpha t}}x = 0\). That yields,\(m = 4,k = 1,\alpha = 0.1\).

Hence, the general solution becomes,

\(\begin{array}{c}x(t) = {c_1}{J_0}\left( {\frac{2}{{0.1}}\sqrt {\frac{1}{4}} {e^{ - 0.1t/2}}} \right) + {c_2}{Y_0}\left( {\frac{2}{{0.1}}\sqrt {\frac{1}{4}} {e^{ - 0.1t/2}}} \right)\\ = {c_1}{J_0}\left( {10{e^{ - t/20}}} \right) + {c_2}{Y_0}\left( {10{e^{ - t/20}}} \right)\end{array}\)

Find the initial value.

Apply the initial conditions.

\(x(0) = {c_1}{J_0}\left( {10{e^0}} \right) + {c_2}{Y_0}\left( {10{e^0}} \right)\)

\(1 = {c_1}{J_0}(10) + {c_2}{Y_0}(10)\)… (1)

\(\begin{array}{c}x'(t) = \frac{d}{{dx}}\left( {{c_1}{J_0}\left( {10{e^{ - t/20}}} \right)} \right) + \frac{d}{{dx}}\left( {{c_2}{Y_0}\left( {10{e^{ - t/20}}} \right)} \right)\\ = \left( {{c_1}J_0^'\left( {10{e^{ - t/20}}} \right) \times \frac{{ - 1}}{{20}} \times 10} \right) + \left( {{c_2}Y_0^'\left( {10{e^{ - t/20}}} \right) \times \frac{{ - 1}}{{20}} \times 10} \right)\\ = - 0.5{c_1}J_0^'\left( {10{e^{ - t/20}}} \right) - 0.5{c_2}Y_0^'\left( {10{e^{ - t/20}}} \right)\end{array}\)

\(\begin{array}{c}x'(0) = - 0.5{c_1}J_0^'\left( {10{e^0}} \right) - 0.5{c_2}Y_0^'\left( {10{e^0}} \right)\\ - \frac{1}{2} = - 0.5{c_1}J_0^'\left( {10} \right) - 0.5{c_2}Y_0^'\left( {10} \right)\\1 = {c_1}J_0^'\left( {10} \right) + {c_2}Y_0^'\left( {10} \right)\end{array}\)

\(1 = {c_1}{J_1}\left( {10} \right) + {c_2}{Y_1}\left( {10} \right)\) … (2)

Find the values using Mathematica.

Solve (1) and (2) using Mathematica that yields,

Hence, the general solution is \(x(t) = - 4.78601{J_0}\left( {10{e^{ - t/20}}} \right) - 3.18028{Y_0}\left( {10{e^{ - t/20}}} \right)\).

Find the graph of the solution obtained using CAS.

Let the graph using Mathematica be,