Internal problem ID [5942]
Internal file name [OUTPUT/5190_Sunday_June_05_2022_03_27_07_PM_65145226/index.tex
]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 1. Introduction– Linear equations of First Order. Page 45
Problem number: 14(b).
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]
\[ \boxed {y^{\prime }-y^{2}=1} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= y^{2}+1 \end {align*}
The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+1}d y &= \int {dx}\\ \arctan \left (y \right )&= x +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \arctan \left (y \right ) = x \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \arctan \left (y\right ) &= x \\
\end{align*} Verification of solutions
\[
\arctan \left (y\right ) = x
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=1, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1+y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{1+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{1+y^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arctan \left (y\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\tan \left (c_{1} \right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\tan \left (x \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\tan \left (x \right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 6
\[
y \left (x \right ) = \tan \left (x \right )
\]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 7
\[
y(x)\to \tan (x)
\]
3.10.2 Solving as quadrature ode
3.10.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([diff(y(x),x)=1+y(x)^2,y(0) = 0],y(x), singsol=all)
DSolve[{y'[x]==1+y[x]^2,{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]