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 \({y_p}(x)\).

8. \(y'' + 3y' - 10y = {x^2}\)

Short answer

\(y = {c_1}{e^{2x}} + {c_2}x{e^{ - 5x}} + \frac{1}{7}\int\limits_{{x_0}}^x {} {t^2}\left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)dt\)

Step-by-step solution

Given Information.

The given problem is

\(y'' + 3y' - 10y = {x^2}\)

Use Differentiate

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

Auxiliary equation corresponding to above equation

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

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

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

Therefore, the solution is

\({y_c} = {c_1}{e^{2x}} + {c_2}{e^{ - 5x}}\)

Wronskian equation

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

Substitute the function values in the differential equation as

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

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

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

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

Apply Green’s function

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

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

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

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

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

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

To find the particular solution

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

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

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

\(\begin{aligned}{c}y = {y_c} + {y_p}\\ = {c_1}{e^{2x}} + {c_2}x{e^{ - 5x}} + \frac{1}{7}\int\limits_{{x_0}}^x {{t^2}} \left( {{e^{2(x - t)}} - {e^{5(t - x)}}} \right)dt\end{aligned}\)

Thus, the required answer is

\(y = {c_1}{e^{2x}} + {c_2}x{e^{ - 5x}} + \frac{1}{7}\int\limits_{{x_0}}^x {} {t^2}\left( {{e^{5(x - t)}} - {e^{2(t - x)}}} \right)dt\)