Internal problem ID [48]
Internal file name [OUTPUT/48_Sunday_June_05_2022_01_34_03_AM_42337791/index.tex]
Book: Differential equations and linear algebra, 3rd ed., Edwards and Penney
Section: Section 1.4. Separable equations. Page 43
Problem number: 23.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {y^{\prime }-2 y=-1} \] With initial conditions \begin {align*} [y \left (1\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-2\\ q(x) &=-1 \end {align*}
Hence the ode is \begin {align*} y^{\prime }-2 y = -1 \end {align*}
The domain of \(p(x)=-2\) is \[
\{-\infty
Integrating both sides gives \begin {align*} \int \frac {1}{-1+2 y}d y &= \int {dx}\\ \frac {\ln \left (-\frac {1}{2}+y \right )}{2}&= x +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} -\frac {\ln \left (2\right )}{2} = 1+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -1-\frac {\ln \left (2\right )}{2} \end {align*}
Trying the constant \begin {align*} c_{1} = -1-\frac {\ln \left (2\right )}{2} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (-\frac {1}{2}+y \right )}{2} = x -1-\frac {\ln \left (2\right )}{2} \end {align*}
The constant \(c_{1} = -1-\frac {\ln \left (2\right )}{2}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (-1+2 y\right )}{2} &= x -1-\frac {\ln \left (2\right )}{2} \\
\end{align*} Verification of solutions
\[
-\frac {\ln \left (2\right )}{2}+\frac {\ln \left (-1+2 y\right )}{2} = x -1-\frac {\ln \left (2\right )}{2}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=-1, y \left (1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-1+2 y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-1+2 y}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-1+2 y}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (-1+2 y\right )}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{2 x +2 c_{1}}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\frac {{\mathrm e}^{2+2 c_{1}}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{2 x -2}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{2 x -2}}{2}+\frac {1}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 14
\[
y \left (x \right ) = \frac {1}{2}+\frac {{\mathrm e}^{2 x -2}}{2}
\]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 18
\[
y(x)\to \frac {1}{2} \left (e^{2 x-2}+1\right )
\]
3.22.2 Solving as quadrature ode
3.22.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([1+diff(y(x),x) = 2*y(x),y(1) = 1],y(x), singsol=all)
DSolve[{1+y'[x] == 2*y[x],y[1]==1},y[x],x,IncludeSingularSolutions -> True]