d 2 y d x 2 + a 2 y = sec ( a x ) The solution to the above equation is y = y c + y p where y c is the complementary factor and y p is the particular integral The auxiliary equation is m 2 + a 2 = 0 ( m − j a ) ( m + j a ) = 0 ∴ m = ± j a { where j is a complex number } Recall that if the solution of the auxiliary equation of a second-order differential equation is m = α ± j β , the general solution is y = e α x ( C 1 cos β x + C 2 sin β x ) ∴ y c = C 1 cos ( a x ) + C 2 sin ( a x ) The Wronskian of the two solutions is W ( x ) = ∣ cos ( a x ) sin ( a x ) d d x ( cos ( a x ) ) d d x ( sin ( a x ) ) ∣ = ∣ cos ( a x ) sin ( a x ) − a sin ( a x ) a cos ( a x ) ∣ = a cos 2 ( a x ) + a sin 2 ( a x ) = a ( cos 2 ( a x ) + sin 2 ( a x ) ) = a ∴ Our particular solution will be given by y p = V 1 ( x ) cos ( a x ) + V 2 ( x ) sin ( a x ) Where V 1 ( x ) = − ∫ r ( x ) sin ( a x ) W ( x ) d x , V 2 ( x ) = ∫ r ( x ) cos ( a x ) W ( x ) d x and V 1 ( x ) = − ∫ r ( x ) sin ( a x ) W ( x ) d x = − ∫ sec ( a x ) sin ( a x ) a d x = − ∫ tan ( a x ) a d x = − ln ( sec ( a x ) ) a 2 + C V 2 ( x ) = ∫ r ( x ) cos ( a x ) W ( x ) d x = ∫ sec ( a x ) cos ( a x ) a d x = ∫ 1 a d x = x a + C The constant terms of the integration can be ignored since we are trying to find a non-constant solution to the differential equation ∴ y = y c + y p = C 1 cos ( a x ) + C 2 sin ( a x ) − 1 a 2 ln ( sec ( a x ) ) cos ( a x ) + x a sin ( a x ) \displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + a^2y = \sec(ax)
\\ \textsf{The solution to the above equation is}
\\ y = y_c + y_p \\
\textsf{where}\, y_c \, \textsf{is the complementary factor and}
\\ y_p\, \textsf{ is the particular integral}
\\ \textsf{The auxiliary equation is} \,m^2 + a^2 = 0 \\
\begin{aligned}
(m - ja)(m + ja) &= 0 \\
\therefore m &= \pm ja \hspace{1cm} \{\textsf{where}\, j\, \textsf{is a complex number}\}
\end{aligned} \\
\textsf{Recall that if the solution of the auxiliary equation}\\
\textsf{of a second-order differential equation is}\\
m = \alpha \pm j\beta, \textsf{the general solution is} \\
y = e^{\alpha x}(C_1\cos{\beta x} + C_2\sin{\beta x})\\
\therefore y_c = C_1\cos(ax) + C_2\sin(ax)\\
\textsf{The Wronskian of the two solutions is} \\
W(x) = \begin{vmatrix}
\cos(ax) & \sin(ax)\\
\frac{\mathrm{d}}{\mathrm{d}x}(\cos(ax)) & \frac{\mathrm{d}}{\mathrm{d}x}(\sin(ax))
\end{vmatrix} =
\begin{vmatrix}
\cos(ax) & \sin(ax)\\
-a\sin(ax) & a\cos(ax)
\end{vmatrix} \\
\begin{aligned}
&= a\cos^2(ax) + a\sin^2(ax)\\
&= a(\cos^2(ax) + \sin^2(ax)) \\
&= a
\end{aligned}\\
\therefore\textsf{ Our particular solution will be given by}\\
y_p = V_1(x)\cos(ax) + V_2(x)\sin(ax)\\
\textsf{Where}\, V_1(x) = -\int \frac{r(x)\sin(ax)}{W(x)} \, \mathrm{d}x,\, V_2(x) = \int \frac{r(x)\cos(ax)}{W(x)}\, \mathrm{d}x \, \textsf{and} \, \\
\begin{aligned}
V_1(x) &= -\int \frac{r(x)\sin(ax)}{W(x)} \, \mathrm{d}x \\
&= -\int \frac{\sec(ax)\sin(ax)}{a}\, \mathrm{d}x\\
&= -\int \frac{\tan(ax)}{a}\, \mathrm{d}x\\
&= -\frac{\ln(\sec(ax))}{a^2} + C
\end{aligned} \\
\begin{aligned}
V_2(x) &= \int \frac{r(x)\cos(ax)}{W(x)}\, \mathrm{d}x \\
&= \int \frac{\sec(ax)\cos(ax)}{a}\, \mathrm{d}x\\
&= \int \frac{1}{a}\, \mathrm{d}x\\
&= \frac{x}{a} + C
\end{aligned} \\
\textsf{The constant terms of the integration can be}
\\\textsf{ignored since we are trying to find a non-constant}
\\\textsf{solution to the differential equation}
\\ \begin{aligned}
\therefore y &= y_c + y_p \\
&=C_1\cos(ax) + C_2\sin(ax) - \frac{1}{a^2}\ln(\sec(ax))\cos(ax) +\frac{x}{a}\sin(ax)
\end{aligned} d x 2 d 2 y + a 2 y = sec ( a x ) The solution to the above equation is y = y c + y p where y c is the complementary factor and y p is the particular integral The auxiliary equation is m 2 + a 2 = 0 ( m − ja ) ( m + ja ) ∴ m = 0 = ± ja { where j is a complex number } Recall that if the solution of the auxiliary equation of a second-order differential equation is m = α ± j β , the general solution is y = e αx ( C 1 cos β x + C 2 sin β x ) ∴ y c = C 1 cos ( a x ) + C 2 sin ( a x ) The Wronskian of the two solutions is W ( x ) = ∣ ∣ cos ( a x ) d x d ( cos ( a x )) sin ( a x ) d x d ( sin ( a x )) ∣ ∣ = ∣ ∣ cos ( a x ) − a sin ( a x ) sin ( a x ) a cos ( a x ) ∣ ∣ = a cos 2 ( a x ) + a sin 2 ( a x ) = a ( cos 2 ( a x ) + sin 2 ( a x )) = a ∴ Our particular solution will be given by y p = V 1 ( x ) cos ( a x ) + V 2 ( x ) sin ( a x ) Where V 1 ( x ) = − ∫ W ( x ) r ( x ) sin ( a x ) d x , V 2 ( x ) = ∫ W ( x ) r ( x ) cos ( a x ) d x and V 1 ( x ) = − ∫ W ( x ) r ( x ) sin ( a x ) d x = − ∫ a sec ( a x ) sin ( a x ) d x = − ∫ a tan ( a x ) d x = − a 2 ln ( sec ( a x )) + C V 2 ( x ) = ∫ W ( x ) r ( x ) cos ( a x ) d x = ∫ a sec ( a x ) cos ( a x ) d x = ∫ a 1 d x = a x + C The constant terms of the integration can be ignored since we are trying to find a non-constant solution to the differential equation ∴ y = y c + y p = C 1 cos ( a x ) + C 2 sin ( a x ) − a 2 1 ln ( sec ( a x )) cos ( a x ) + a x sin ( a x )
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