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).

\(y'' + 3y' - 10y = f(x)\)

Short answer

\({y_p}(x) = \frac{1}{7}\int\limits_{{x_0}}^x {} \left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)f(t)dt\)

Step-by-step solution

Step 1:Given Information.

The given problem is

\(y'' + 3y' - 10y = f(x)\)

Use Differentiate

\(y'' + 3y' - 10y = 0\)

Assume as

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

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

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

Differentiate we get

\({m^2}{e^{mx}} + 3m{e^{mx}} - 10{e^{mx}} = 0\)

\({e^{mx}}\left( {{m^2} + 3m - 10} \right) = 0\)

\({m^2} + 3m - 10 = 0\)

\((m - 2)(m + 5) = 0\)

\({m_1} = 2\;and\;{m_2} = - 5\)

\({y_1} = {e^{2x}}\;and\;{y_2} = {e^{ - 5x}}\)

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

\( = \left| {\begin{aligned}{*{20}{c}}{{e^{2x}}}&{{e^{ - 5x}}}\\{2{e^{2x}}}&{ - 5{e^{ - 5x}}}\end{aligned}} \right|\)

\( = \left( {{e^{2x}}} \right) \times \left( { - 5{e^{ - 5x}}} \right) - \left( {{e^{ - 5x}}} \right) \times \left( {2{e^{2x}}} \right)\)

\( = - 5{e^{ - 3x}} - 2{e^{ - 3x}}\)

\( = - 7{e^{ - 3x}}\)

Apply Green’s function

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

\( = \frac{{{e^{2t}} \times {e^{ - 5x}} - {e^{2x}} \times {e^{ - 5t}}}}{{ - 7{e^{ - 3x}}}}\)

\( = \frac{{{e^{(2t - 5x)}} - {e^{(2x - 5t)}}}}{{ - 7{e^{ - 3x}}}}\)

\( = \frac{{{e^{(2x - 5t)}} - {e^{(2t - 5x)}}}}{{7{e^{ - 3x}}}} \times \frac{{{e^{3x}}}}{{{e^{3x}}}}\)

\( = \frac{{{e^{(5x - 5t)}} - {e^{(2t - 2x)}}}}{7}\)

\( = \frac{1}{7}\left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)\)

To find the particular solution

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

\( = \int\limits_{{x_0}}^x {} \frac{1}{7}\left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)f(t)dt\)

\( = \frac{1}{7}\int\limits_{{x_0}}^x {} \left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)f(t)dt\)

\({y_p}(x) = \frac{1}{7}\int\limits_{{x_0}}^x {} \left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)f(t)dt\)