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).

6. \(y'' - 2y' + 2y = f(x)\)

Short answer

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

Step-by-step solution

Given Information.

The given problem is

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

Use Differentiate

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

Auxiliary equation corresponding to the above equation

\(\left( {{m^2} - 2m + 2} \right) = 0\)

\(\begin{aligned}{c}{m_{1,2}} = \frac{{2 \pm \sqrt {4 - 4 \times 1 \times 2} }}{2}\\ = \frac{{2 \pm \sqrt { - 4} }}{2}\\ = 1 \pm i\end{aligned}\)

Therefore the solution is

\({y_c} = {e^x}({C_1}{\mathop{\rm Cos}\nolimits} x + {C_2}{\mathop{\rm Sin}\nolimits} x)\)

Wronskian equation

\(\begin{aligned}{c}W(t) = W\left( {{y_1},{y_2}} \right)\\ = \left| {\begin{aligned}{*{20}{c}}{{y_1} {y_2}}\\{y_1' y_2'}\end{aligned}} \right|\end{aligned}\)

Substitute the function values in the differential equation as

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

\( = \left( {{e^t}cost} \right) \times \left( {{e^t}cost + {e^t}sint} \right) - \left( {{e^t}sint} \right) \times \left( {{e^t}cost - {e^t}sint} \right)\)

\( = \left( {{e^{2t}}co{s^2}t + {e^{2t}}sintcost} \right) - \left( {{e^{2t}}sintcost - {e^{2t}}si{n^2}t} \right)\)

\( = {e^{2t}}\left( {co{s^2}t + si{n^2}t} \right) + {e^{2t}}(sintcost - sintcost)\)

\( = {e^{2t}}\)

Apply Green’s function

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

\( = \frac{{{e^t}cost \times {e^x}sinx - {e^x}cosx \times {e^t}sint}}{{{e^{2t}}}}\)

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

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

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

\( = {e^{(x - t)}}sin(x - t)\)

\( = {e^{(x - t)}}sin(x - t)\)

To find the particular solution

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

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

Thus, the required answer is

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