5.41 problem 1576

Internal problem ID [9898]
Internal file name [OUTPUT/8845_Monday_June_06_2022_05_37_12_AM_79064382/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1576.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"

Maple gives the following as the ode type

[[_high_order, _missing_x]]

\[ \boxed {f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )+2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )=0} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} = 0 \] The characteristic equation is \[ a^{4} f -2 a^{2} f \,\lambda ^{2}+f \,\lambda ^{4}-2 a^{2} \operatorname {df} \lambda +2 \operatorname {df} \,\lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= a\\ \lambda _2 &= -a\\ \lambda _3 &= \frac {-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f}\\ \lambda _4 &= -\frac {\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{1} +{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{2} +{\mathrm e}^{x a} c_{3} +{\mathrm e}^{-x a} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \\ y_2 &= {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \\ y_3 &= {\mathrm e}^{x a} \\ y_4 &= {\mathrm e}^{-x a} \\ \end{align*} Now the particular solution to the given ODE is found \[ y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} = -f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coefficient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3&y_4\\ y_1'&y_2'&y_3'&y_4'\\ y_1''&y_2''&y_3''&y_4''\\ y_1'''&y_2'''&y_3'''&y_4'''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{cccc} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{x a} & {\mathrm e}^{-x a} \\ -\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & a \,{\mathrm e}^{x a} & -a \,{\mathrm e}^{-x a} \\ \frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & a^{2} {\mathrm e}^{x a} & {\mathrm e}^{-x a} a^{2} \\ -\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{3} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{3}} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{3} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{3}} & a^{3} {\mathrm e}^{x a} & -a^{3} {\mathrm e}^{-x a} \end {array}\right ] \\ |W| &= \frac {16 \,{\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} a \,{\mathrm e}^{x a} {\mathrm e}^{-x a} \operatorname {df}^{2} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\, \left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{5}} \end {align*}

The determinant simplifies to \begin {align*} |W| &= \frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{3}} \end {align*}

Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{x a} & {\mathrm e}^{-x a} \\ \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & a \,{\mathrm e}^{x a} & -a \,{\mathrm e}^{-x a} \\ \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & a^{2} {\mathrm e}^{x a} & {\mathrm e}^{-x a} a^{2} \end {array}\right ] \\ &= \frac {4 a \operatorname {df} \,{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}} \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{x a} & {\mathrm e}^{-x a} \\ -\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & a \,{\mathrm e}^{x a} & -a \,{\mathrm e}^{-x a} \\ \frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & a^{2} {\mathrm e}^{x a} & {\mathrm e}^{-x a} a^{2} \end {array}\right ] \\ &= -\frac {4 a \operatorname {df} \,{\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}} \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{-x a} \\ -\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & -a \,{\mathrm e}^{-x a} \\ \frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & {\mathrm e}^{-x a} a^{2} \end {array}\right ] \\ &= -\frac {4 a \operatorname {df} \,{\mathrm e}^{-\frac {x \left (f a +2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}} \end {align*}

\begin {align*} W_4(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} & {\mathrm e}^{x a} \\ -\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f} & a \,{\mathrm e}^{x a} \\ \frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & \frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )^{2} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}}{f^{2}} & a^{2} {\mathrm e}^{x a} \end {array}\right ] \\ &= \frac {4 a \operatorname {df} \,{\mathrm e}^{\frac {x \left (f a -2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}} \end {align*}

Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{4-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{3} \int { \frac { \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df}\right ) \left (\frac {4 a \operatorname {df} \,{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}}\right )}{\left (f\right ) \left (\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{3}}\right )} \, dx} \\ &= - \int { \frac {\frac {4 \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} \right ) a \operatorname {df} \,{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}}}{\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}} \, dx}\\ &= - \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

\begin {align*} U_2 &= (-1)^{4-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df}\right ) \left (-\frac {4 a \operatorname {df} \,{\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}}\right )}{\left (f\right ) \left (\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{3}}\right )} \, dx} \\ &= \int { \frac {-\frac {4 \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} \right ) a \operatorname {df} \,{\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right )}{f^{2}}}{\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}} \, dx}\\ &= \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

\begin {align*} U_3 &= (-1)^{4-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df}\right ) \left (-\frac {4 a \operatorname {df} \,{\mathrm e}^{-\frac {x \left (f a +2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}\right )}{\left (f\right ) \left (\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{3}}\right )} \, dx} \\ &= - \int { \frac {-\frac {4 \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} \right ) a \operatorname {df} \,{\mathrm e}^{-\frac {x \left (f a +2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}}{\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}} \, dx}\\ &= - \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

\begin {align*} U_4 &= (-1)^{4-4} \int { \frac {F(x) W_4(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df}\right ) \left (\frac {4 a \operatorname {df} \,{\mathrm e}^{\frac {x \left (f a -2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}\right )}{\left (f\right ) \left (\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{3}}\right )} \, dx} \\ &= \int { \frac {\frac {4 \left (-f \left (y^{\prime \prime \prime \prime }-2 a^{2} y^{\prime \prime }+a^{4} y\right )-2 \operatorname {df} \left (y^{\prime \prime \prime }-a^{2} y^{\prime }\right )+y a^{4} f -2 y^{\prime \prime } a^{2} f -2 y^{\prime } a^{2} \operatorname {df} +y^{\prime \prime \prime \prime } f +2 y^{\prime \prime \prime } \operatorname {df} \right ) a \operatorname {df} \,{\mathrm e}^{\frac {x \left (f a -2 \operatorname {df} \right )}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}}{\frac {16 a^{3} \operatorname {df}^{2} {\mathrm e}^{-\frac {2 x \operatorname {df}}{f}} \sqrt {a^{2} f^{2}+\operatorname {df}^{2}}}{f^{2}}} \, dx}\\ &= \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Hence \begin {equation*} \begin {split} y_p &= \left (0\right ) \left ({\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}\right ) \\ &+\left (0\right ) \left ({\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}\right ) \\ &+\left (0\right ) \left ({\mathrm e}^{x a}\right ) \\ &+\left (0\right ) \left ({\mathrm e}^{-x a}\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = 0 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{1} +{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{2} +{\mathrm e}^{x a} c_{3} +{\mathrm e}^{-x a} c_{4}\right ) + \left (0\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{1} +{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{2} +{\mathrm e}^{x a} c_{3} +{\mathrm e}^{-x a} c_{4} \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{1} +{\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} c_{2} +{\mathrm e}^{x a} c_{3} +{\mathrm e}^{-x a} c_{4} \] Verified OK.

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 67

dsolve(f*(diff(y(x),x$4)-2*a^2*diff(y(x),x$2)+a^4*y(x))+2*df*(diff(y(x),x$3)-a^2*diff(y(x),x))=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} {\mathrm e}^{a x}+c_{2} {\mathrm e}^{-a x}+c_{3} {\mathrm e}^{\frac {\left (-\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}}+c_{4} {\mathrm e}^{-\frac {\left (\operatorname {df} +\sqrt {a^{2} f^{2}+\operatorname {df}^{2}}\right ) x}{f}} \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 80

DSolve[f*(y''''[x]-2*a^2*y''[x]+a^4*y[x])+2*df*(y'''[x]-a^2*y'[x])==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 e^{\frac {x \left (\sqrt {a^2 f^2+\text {df}^2}-\text {df}\right )}{f}}+c_2 e^{-\frac {x \left (\sqrt {a^2 f^2+\text {df}^2}+\text {df}\right )}{f}}+c_3 e^{-a x}+c_4 e^{a x} \]