Question

In Problems 7–12 proceed as in Example 2 to fi­nd the general solution of the given differential equation. Use the results obtained in Problems 1–6. Do not evaluate the integral that de­fines \({y_p}(x)\).

7. \(y'' - 16y = x{e^{ - 2x}}\)

Short answer

\(y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}} + \frac{1}{4}\int\limits_{{x_o}}^x {} \frac{{tsinh4(x - t)}}{{{e^{2t}}}}dt\)

Step-by-step solution

Given Information.

The given problem is

\(y'' - 16y = x{e^{ - 2x}}\)

Use Differentiate

\(y'' - 16y = 0\)

Auxiliary equation corresponding to above equation

\({m^2} - 16 = 0\)

\((m - 4)(m + 4) = 0\)

\({m_1} = 4\;and\;{m_2} = - 4\)

Therefore, the solution is

\({y_c} = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}}\)

Wronskian equation

\(\begin{aligned}{c}W(t) = W\left( {{y_1},{y_2}} \right)\\ = \left| {\begin{aligned}{*{20}{c}}{{y_1} {y_2}}\\{y_1' y_2'}\end{aligned}} \right|\end{aligned}\)

Substitute the function values in the differential equation as

\( = \left| {\begin{aligned}{*{20}{c}}{{e^{4t}}}&{{e^{ - 4t}}}\\{4{e^{4t}}}&{ - 4{e^{ - 4t}}}\end{aligned}} \right|\)

\( = \left( {{e^{4t}}} \right) \times \left( { - 4{e^{ - 4t}}} \right) - \left( {{e^{ - 4t}}} \right) \times \left( {4{e^{4t}}} \right)\)

\( = - 4{e^0} - 4{e^0}\)

\( = - 4 - 4\)

\( = - 8\)

Apply Green’s function

\(G(x,t) = \frac{{{y_1}(t){y_2}(x) - {y_1}(x){y_2}(t)}}{{W\left( t \right)}}\)

\( = \frac{{{e^{4t}} \times {e^{ - 4x}} - {e^{4x}} \times {e^{ - 4t}}}}{{ - 8}}\)

\( = \frac{{{e^{(4t - 4x)}} - {e^{(4x - 4t)}}}}{{ - 8}}\)

\( = \frac{{{e^{(4x - 4t)}} - {e^{(4t - 4x)}}}}{8}\)

\( = \frac{{{e^{4(x - t)}} - {e^{4(t - x)}}}}{8}\)

\( = \frac{1}{4}\frac{{{e^{4(x - t)}} - {e^{4(t - x)}}}}{2}\)

\( = \frac{1}{4}sinh4(x - t)\)

To find the particular solution

\({y_p}(x) = \int\limits_{{x_0}}^x {} G(x,t)f(t)dt\)

\( = \int\limits_{{x_0}}^x {} \frac{1}{4}sinh4(x - t)t{e^{ - 2t}}dt\)

\( = \frac{1}{4}\int\limits_{{x_0}}^x {} t{e^{ - 2t}}sinh4(x - t)dt\)

\(\begin{aligned}{c}y = {y_c} + {y_p}\\ = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}} + \frac{1}{4}\int\limits_{{x_0}}^x {} t{e^{ - 2t}}sinh4(x - t)dt\\ = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}} + \frac{1}{4}\int\limits_{{x_o}}^x {} \frac{{tsinh4(x - t)}}{{{e^{2t}}}}dt\end{aligned}\)

Thus the required answer is

\(y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}} + \frac{1}{4}\int\limits_{{x_o}}^x {} \frac{{tsinh4(x - t)}}{{{e^{2t}}}}dt\)