Internal problem ID [13549]
Internal file name [OUTPUT/12721_Friday_February_16_2024_12_11_13_AM_73807896/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 14. Higher order equations and the reduction of order method. Additional
exercises page 277
Problem number: 14.3 (a).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-4 y^{\prime }+3 y=9 \,{\mathrm e}^{2 x}} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{3 x} \end {align*}
This is second order nonhomogeneous ODE. In standard form the ODE is \[ A y''(x) + B y'(x) + C y(x) = f(x) \] Where \(A=1, B=-4, C=3, f(x)=9 \,{\mathrm e}^{2 x}\). Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the inhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ y^{\prime \prime }-4 y^{\prime }+3 y = 0 \] Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = -4 \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{3 x} \left (\int {\mathrm e}^{-\left (\int \left (-4\right )d x \right )} {\mathrm e}^{-6 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{3 x} \int \frac {{\mathrm e}^{4 x}}{{\mathrm e}^{6 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{3 x} \left (\int {\mathrm e}^{-2 x}d x \right ) \\ y_{2}\left (x \right ) &= -\frac {{\mathrm e}^{3 x} {\mathrm e}^{-2 x}}{2} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{3 x} {\mathrm e}^{-2 x}}{2} \\ \end{align*} Therefore the homogeneous solution \(y_h\) is \[ y_h = c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{3 x} {\mathrm e}^{-2 x}}{2} \] The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 9 \,{\mathrm e}^{2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{-\frac {{\mathrm e}^{3 x} {\mathrm e}^{-2 x}}{2}, {\mathrm e}^{3 x}\right \} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} {\mathrm e}^{2 x} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ -A_{1} {\mathrm e}^{2 x} = 9 \,{\mathrm e}^{2 x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = -9] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -9 \,{\mathrm e}^{2 x} \] Therefore the general solution is \begin {align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{3 x} {\mathrm e}^{-2 x}}{2}\right ) + \left (-9 \,{\mathrm e}^{2 x}\right ) \end {align*}
Which simplifies to \[ y = c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{x}}{2}-9 \,{\mathrm e}^{2 x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{x}}{2}-9 \,{\mathrm e}^{2 x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{3 x}-\frac {c_{2} {\mathrm e}^{x}}{2}-9 \,{\mathrm e}^{2 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-4 y^{\prime }+3 y=9 \,{\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +3=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1, 3\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{3 x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{3 x}+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=9 \,{\mathrm e}^{2 x}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{x} & {\mathrm e}^{3 x} \\ {\mathrm e}^{x} & 3 \,{\mathrm e}^{3 x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=2 \,{\mathrm e}^{4 x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {9 \,{\mathrm e}^{x} \left (\int {\mathrm e}^{x}d x \right )}{2}+\frac {9 \,{\mathrm e}^{3 x} \left (\int {\mathrm e}^{-x}d x \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-9 \,{\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{3 x}-9 \,{\mathrm e}^{2 x} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 21
dsolve([diff(y(x),x$2)-4*diff(y(x),x)+3*y(x)=9*exp(2*x),exp(3*x)],singsol=all)
\[ y \left (x \right ) = c_{2} {\mathrm e}^{3 x}+c_{1} {\mathrm e}^{x}-9 \,{\mathrm e}^{2 x} \]
✓ Solution by Mathematica
Time used: 0.069 (sec). Leaf size: 25
DSolve[y''[x]-4*y'[x]+3*y[x]==9*Exp[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (-9 e^x+c_2 e^{2 x}+c_1\right ) \]