Internal problem ID [12868]
Internal file name [OUTPUT/11521_Monday_November_06_2023_01_31_17_PM_12857361/index.tex
]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.2. page
33
Problem number: 8.
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]
\[ \boxed {y^{\prime }+y=2} \]
Integrating both sides gives \begin {align*} \int \frac {1}{-y +2}d y &= t +c_{1}\\ -\ln \left (y -2\right )&=t +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{-t -c_{1}}+2\\ &=\frac {{\mathrm e}^{-t}}{c_{1}}+2 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-t}}{c_{1}}+2 \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-t}}{c_{1}}+2 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y=2 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2-y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2-y}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{2-y}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (2-y\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-{\mathrm e}^{-t -c_{1}}+2 \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
dsolve(diff(y(t),t)=2-y(t),y(t), singsol=all)
\[ y \left (t \right ) = 2+{\mathrm e}^{-t} c_{1} \]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 20
DSolve[y'[t]==2-y[t],y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to 2+c_1 e^{-t} \\ y(t)\to 2 \\ \end{align*}