Internal problem ID [279]
Internal file name [OUTPUT/279_Sunday_June_05_2022_01_38_11_AM_2821455/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 5.2, Higher-Order Linear Differential Equations. General solutions of Linear
Equations. Page 288
Problem number: problem 39.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeff", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {4 y^{\prime \prime }-4 y^{\prime }+y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{\frac {x}{2}} \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 \[ p \left (x \right ) = -1 \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{\frac {x}{2}} \left (\int {\mathrm e}^{-\left (\int \left (-1\right )d x \right )} {\mathrm e}^{-x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{\frac {x}{2}} \int \frac {{\mathrm e}^{x}}{{\mathrm e}^{x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{\frac {x}{2}} \left (\int 1d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{\frac {x}{2}} x \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= {\mathrm e}^{\frac {x}{2}} c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}} x \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\frac {x}{2}} c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}} x \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\frac {x}{2}} c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime }-4 y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=y^{\prime }-\frac {y}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-y^{\prime }+\frac {y}{4}=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-r +\frac {1}{4}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (2 r -1\right )^{2}}{4}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\frac {1}{2} \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{\frac {x}{2}} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{\frac {x}{2}} x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{\frac {x}{2}} c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}} x \end {array} \]
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: 0.0 (sec). Leaf size: 14
dsolve([4*diff(y(x),x$2)-4*diff(y(x),x)+y(x)=0,exp(x/2)],singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {x}{2}} \left (c_{2} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 20
DSolve[4*y''[x]-4*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{x/2} (c_2 x+c_1) \]