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).

\(y'' + 9y = f(x)\)

Short answer

\({y_p}(x) = \frac{1}{3}\int\limits_{{x_0}}^x {} sin(3(x - t))f(t)dt\)

Step-by-step solution

Given Information.

The given problem is

\(y'' + 9y = f(x)\)

Use Differentiate

\(y'' + 9y = 0\)

Assume as

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

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

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

Differentiate we get

\({m^2}{e^{mx}} + 9{e^{mx}} = 0\)

\({e^{mx}}\left( {{m^2} + 9} \right) = 0\)

\({m^2} + 9 = 0\)

\({m^2} = - 9\)

\(m = \pm \sqrt 9 \)

\(m = \pm 3i\)

\({y_1} = cos3x\;and\;{y_2} = sin3x\)

Wronskian equation

Substitute the function values in the differential equation as

\( = (cos3x) \times (3cos3x) - (sin3x) \times ( - 3sin3x)\)

\( = 3co{s^2}3x - \left( { - 3si{n^2}3x} \right)\)

\( = 3\left( {co{s^2}3x + si{n^2}3x} \right)\)

\( = 3\)

Apply Green’s function

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

\( = \frac{{cos3t \times sin3x - cos3x \times sin3t}}{3}\)

\( = \frac{{sin3xcos3t - cos3xsin3t}}{3}\)

\( = \frac{{sin(3(x - t))}}{3}\)

\( = \frac{1}{3}sin(3(x - t))\)

\(sin(3(x - t)) = sin3xcos3t - cos3xsin3t\)

To find the particular solution

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

\( = \int\limits_{{x_0}}^x {} \frac{1}{3}sin(3(x - t))f(t)dt\)

\( = \frac{1}{3}\int\limits_{{x_0}}^x {} sin(3(x - t))f(t)dt\)

\({y_p}(x) = \frac{1}{3}\int\limits_{{x_0}}^x {} sin(3(x - t))f(t)dt\)