Internal problem ID [13269]
Internal file name [OUTPUT/12441_Wednesday_February_14_2024_02_06_16_AM_19470867/index.tex
]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 2. Integration and differential equations. Additional exercises. page
32
Problem number: 2.4 (f).
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]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime }=1} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \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) &=0\\ q(x) &=\frac {1}{x^{2}+1} \end {align*}
Hence the ode is \begin {align*} y^{\prime } = \frac {1}{x^{2}+1} \end {align*}
The domain of \(p(x)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} y &= \int { \frac {1}{x^{2}+1}\,\mathop {\mathrm {d}x}}\\ &= \arctan \left (x \right )+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=3\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 3 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 3 \end {align*}
Trying the constant \begin {align*} c_{1} = 3 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\arctan \left (x \right )+3 \end {align*}
The constant \(c_{1} = 3\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \arctan \left (x \right )+3 \\
\end{align*} Verification of solutions
\[
y = \arctan \left (x \right )+3
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (x^{2}+1\right ) y^{\prime }=1, y \left (0\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\arctan \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\arctan \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\arctan \left (x \right )+3 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\arctan \left (x \right )+3 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 8
\[
y \left (x \right ) = \arctan \left (x \right )+3
\]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 9
\[
y(x)\to \arctan (x)+3
\]
1.28.2 Solving as quadrature ode
1.28.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([(x^2+1)*diff(y(x),x)=1,y(0) = 3],y(x), singsol=all)
DSolve[{(x^2+1)*y'[x]==1,{y[0]==3}},y[x],x,IncludeSingularSolutions -> True]