Question

In Problems, 1–6 proceed as in Example 1 to and a particular solution

\({y_p}(x)\)of the given differential equation in the integral form (10).

\(y'' + 2y' + y = f(x)\)

Short answer

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

Step-by-step solution

Step 1:Given Information.

The given problem is:

\(y'' + 2y' + y = f(x)\)

Use Differentiate

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

Assume as

\(y = {e^{mx}}\)

\(y' = m{e^{mx}}\)

\(y'' = {m^2}{e^{mx}}\)

Differentiate we get

\({m^2}{e^{mx}} + 2m{e^{mx}} + {e^{mx}} = 0\)

\({e^{mx}}\left( {{m^2} + 2m + 1} \right) = 0\)

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

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

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

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

Wronskian equation

\(W\left( {{y_1},{y_2}} \right) = \left| {\begin{aligned}{*{20}}{{y_1} {y_2}}\\{y_1' y_2'}\end{aligned}} \right|\)

Substitute the function values in the differential equation as

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

\( = \left( {{e^{ - x}}} \right) \times \left( { - x{e^{ - x}} + {e^{ - x}}} \right) - \left( {x{e^{ - x}}} \right) \times \left( { - {e^{ - x}}} \right)\)

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

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

Apply Green’s function

\(\begin{aligned}{c}G(x,t) &= \frac{{{y_1}(t){y_2}(x) - {y_1}(x){y_2}(t)}}{{W\left( {{y_1},{y_2}} \right)}}\\ &= \frac{{{e^{ - t}} \times x{e^{ - x}} - {e^{ - x}} \times t{e^{ - t}}}}{{{e^{ - 2x}}}}\\ &= \frac{{x{e^{ - x - t}} - t{e^{ - x - t}}}}{{{e^{ - 2x}}}}\\ &= \frac{{x{e^{ - x - t}} - t{e^{ - x - t}}}}{{{e^{ - 2x}}}} \times \frac{{{e^{2x}}}}{{{e^{2x}}}}\\ &= \frac{{x{e^{x - t}} - t{e^{x - t}}}}{{{e^{ - 2x}} \times {e^{2x}}}}\\ &= (x - t){e^{x - t}}\end{aligned}\)

To find the particular solution

\(\begin{aligned}{c}{y_p}(x) &= \int\limits_{{x_0}}^x {} G(x,t)f(t)dt\\ &= \int\limits_{{x_0}}^x {} (x - t){e^{x - t}}f(t)dt\\{y_p}(x) &= \int\limits_{{x_0}}^x {} (x - t){e^{x - t}}f(t)dt\end{aligned}\)