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 \({y_p}(x)\).

9. \(y'' + 2y' + y = {e^{ - x}}\)

Short answer

\(y = {c_1}{e^{ - x}} + {c_2}x{e^{ - x}} + \int\limits_{{x_0}}^x {} {e^{ - x + t}}(x - t){e^{ - t}}dt\)

Step-by-step solution

Given Information.

The given problem is

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

Use Differentiate

\(y'' + 2y' + y = 0\)

Auxiliary equation corresponding to above equation

\({m^2} + 2m + 1 = 0\)

\({m_1} = {m_2} = - 1\)

\({m_1} = - 1,{m_2} = - 1\)

Therefore, the solution is

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

\({y_1}(x) = {e^{ - x}}\;and\;{y_2}(x) = x{e^{ - x}}\)

Wronskian equation

\(W(t) = W\left( {{e^{ - t}},t{e^{ - t}}} \right)\)

\(W(t) = \left| {\begin{aligned}{*{20}{c}}{{e^{ - t}}}&{t{e^{ - t}}}\\{ - {e^{ - t}}}&{{e^{ - t}}(1 - t)}\end{aligned}} \right|\)

Substitute the function values in the differential equation as

\( = {e^{ - t}} \times {e^{ - t}}(1 - t) - t{e^{ - t}} \times \left( { - {e^{ - t}}} \right)\)

\( = {e^{ - 2t}}(1 - t) + t{e^{ - 2t}}\)

\( = {e^{ - 2t}}(1 - t + t)\)

\( = {e^{ - 2t}}\)

Apply Green’s function

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

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

\( = {e^t}x{e^{ - x}} - {e^{ - x}}t{e^t}\)

\( = x{e^{ - x + t}} - t{e^{ - x + t}}\)

\( = {e^{ - x + t}}(x - t)\)

To find the particular solution

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

\({y_p} = \int\limits_{{x_0}}^x {} {e^{ - x + t}}(x - t)f(t)dt\)

\({y_p} = \int\limits_{{x_0}}^x {} {e^{ - x + t}}(x - t){e^{ - t}}dt\)

\(y = {y_c} + {y_p}\)

\(y = {c_1}{e^{ - x}} + {c_2}x{e^{ - x}} + \int\limits_{{x_0}}^x {} {e^{ - x + t}}(x - t){e^{ - t}}dt\)