Question

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 \({y_p}(x)\).

11.

Short answer

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Step-by-step solution

Given Information.

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Determining the General solution:

We have a second-order differential equation on our hands.

Then we'll use the following strategy to find a generic solution.

We must first locate the solutions to the associated homogeneous differential equations.

, \(\left( 1 \right)\)

Assume that \(y = {e^{mx}}\)is a solution for this homogeneous differential equation, then differentiate this assumption with respect to x, and you'll get

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

\(\left( 3 \right)\)

After that, we insert \(y = {e^{mx}}\) and equations \(\left( 2 \right)\) and \(\left( 3 \right)\) into differential equation \(\left( 1 \right)\) and get

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

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

We have e because \({e^{mx}}\) cannot equal \(0\).

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

The roots are then

\({m_{1,2}} = \pm 3i\)

which is a conjugate compound.

The solution of the associated homogeneous differential equation is then as follows:

\({y_c} = {c_1} \cos 3x + {c_2}\sin 3x\)

Determining the particular solution \({y_p}\left( x \right)\):

Second, we must use the following technique to obtain the specific solution \({y_p}(x)\) of the above differential equation:

We can get the wronskian \(W\left( {{y_1},{y_2}} \right)\) where, \({y_1} = \cos 3t\)and \({y_2} = \sin 3t\)

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

\( = \left| {\begin{aligned}{*{20}{c}}{cos3t}&{sin3t}\\{ - 3 sin 3t}&{3 cos 3t}\end{aligned}} \right|\)

\( = (cos 3t) \times (3 cos 3t) - (sin 3t) \times ( - 3 sin 3t)\)

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

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

\( = 3\)

Then, for the above differential equation, we must obtain the Green's function as follows:

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

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

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

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

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

where we know that \(sin(x - t) = sin x cos t - cos x sin t\)thanks to trigonometric identities

 Step 4: Using Green’s function:

Then, using the Green's function and \(f\left( t \right) = t + \sin t\), we must obtain the specific solution \({y_p}\left( x \right)\) as

\({y_p}\left( x \right) = \int_{{x_0}}^x {G\left( {x - t} \right)} f\left( t \right) dt\)
\( = \int_{{x_0}}^x {\frac{1}{3}\sin \left( {x - t} \right)} \left( {t + \sin t} \right) dt\)

\( = \frac{1}{3}\int_{{x_0}}^x {\sin \left( {x - t} \right)} \left( {t + \sin t} \right) dt\) \(\left( b \right)\)

Finally, we may derive the general solution of the given differential equation by combining \(\left( a \right)\) and \(\left( b \right)\).

\(y = {y_c} + {y_p}\)

\( = {c_1} \cos 3x + {c_2}\sin 3x + \frac{1}{3}\int_{{x_0}}^x {\sin \left( {x - t} \right)} \left( {t + \sin t} \right) dt\)

Determining the Result:

xy