Internal problem ID [6338]
Internal file name [OUTPUT/5586_Sunday_June_05_2022_03_44_14_PM_46831161/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 2. Second-Order Linear Equations. Section 2.4. THE USE OF A KNOWN
SOLUTION TO FIND ANOTHER. Page 74
Problem number: 1(b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeff", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }-y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{x} \end {align*}
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 \[ p \left (x \right ) = 0 \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int {\mathrm e}^{-\left (\int 0d x \right )} {\mathrm e}^{-2 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \int \frac {1}{{\mathrm e}^{2 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int {\mathrm e}^{-2 x}d x \right ) \\ y_{2}\left (x \right ) &= -\frac {{\mathrm e}^{-2 x} {\mathrm e}^{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}^{x}-\frac {c_{2} {\mathrm e}^{-2 x} {\mathrm e}^{x}}{2} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}-\frac {c_{2} {\mathrm e}^{-2 x} {\mathrm e}^{x}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{x}-\frac {c_{2} {\mathrm e}^{-2 x} {\mathrm e}^{x}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-x} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 15
dsolve([diff(y(x),x$2)-y(x)=0,exp(x)],singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} \]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 20
DSolve[y''[x]-y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 e^x+c_2 e^{-x} \]