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).

\(4y'' - 4y' + y = f(x)\)

Short answer

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

Step-by-step solution

Step 1:Given Information.

The given problem is

\(4y'' - 4y' + y = f(x)\)

Use Differentiate

\(4y'' - 4y' + y = 0\)

Assume as

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

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

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

Differentiate we get

\(4{m^2}{e^{mx}} - 4m{e^{mx}} + {e^{mx}} = 0\)

\({e^{mx}}\left( {4{m^2} - 4m + 1} \right) = 0\)

\(4{m^2} - 4m + 1 = 0\)

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

\(2m = 1,2m = 1\)

\({m_1} = \frac{1}{2},{m_2} = \frac{1}{2}\)

\({y_1} = {e^{\frac{1}{2}x}}\;and\;{y_2} = x{e^{\frac{1}{2}x}}\)

Wronskian equation

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

Substitute the function values in the differential equation as

\( = \left| {\begin{aligned}{e^{\frac{1}{2}x}}&{x{e^{\frac{1}{2}x}}}\\{\frac{1}{2}{e^{\frac{1}{2}x}}}&{\frac{1}{2}x{e^{\frac{1}{2}x}} + {e^{\frac{1}{2}x}}}\end{aligned}} \right|\)

\( = \left( {{e^{\frac{1}{2}x}}} \right) \times \left( {\frac{1}{2}x{e^{\frac{1}{2}x}} + {e^{\frac{1}{2}x}}} \right) - \left( {\frac{1}{2}{e^{\frac{1}{2}x}}} \right) \times \left( {x{e^{\frac{1}{2}x}}} \right)\)

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

\( = {e^x}\)

Apply Green’s function

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

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

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

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

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

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

To find the particular solution

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

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