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).

\(y'' - 16y = f(x)\)

Short answer

\({y_p}(x) = \frac{1}{4}\int_{{x_0}}^x {sinh4(x - t)f(t)dt} \)

Step-by-step solution

Step 1:Given Information.

The given problem is

\(y'' - 16y = f(x)\)

Use Differentiate

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

Assume as

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

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

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

Differentiate we get

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

\({e^{mx}}\left( {{m^2} - 16} \right) = 0\)

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

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

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

\({y_1} = {e^{4x}}\;and\;{y_2} = {e^{ - 4x}}\)

Step 3:Wronskian equation

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

Substitute the function values in the differential equation as

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

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

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

\( = - 4 - 4\)

\( = - 8\)

Apply Green’s function

\(\begin{aligned}{c}G(x,t) &= \frac{{{y_1}(t){y_2}(x) - {y_1}(x){y_2}(t)}}{{W\left( {{y_1},{y_2}} \right)}}\\ &= \frac{{{e^{4t}} \times {e^{ - 4x}} - {e^{4x}} \times {e^{ - 4t}}}}{{ - 8}}\\ &= \frac{{{e^{(4t - 4x)}} - {e^{(4x - 4t)}}}}{{ - 8}}\\ &= \frac{{{e^{4(x - t)}} - {e^{4(t - x)}}}}{8}\\ &= \frac{1}{4}\frac{{{e^{4(x - t)}} - {e^{4(t - x)}}}}{2}\\ &= \frac{1}{4}sinh4(x - t)\end{aligned}\)

To find the particular solution

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

\( = \int_{{x_0}}^x {\frac{1}{4}sinh4(x - t)f(t)dt} \)

\( = \frac{1}{4}\int_{{x_0}}^x {sinh4(x - t)f(t)dt} \)

\({y_p}(x) = \frac{1}{4}\int_{{x_0}}^x {sinh4(x - t)f(t)dt} \)