Internal problem ID [12708]
Internal file name [OUTPUT/11361_Friday_November_03_2023_06_31_05_AM_38277058/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.4.4, page 115
Problem number: 6 (a).
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]
\[ \boxed {y^{\prime }-y^{3}=0} \] With initial conditions \begin {align*} [y \left (-1\right ) = 1] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= y^{3} \end {align*}
The \(y\) domain of \(f(x,y)\) when \(x=-1\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=-1\) is \[
\{-\infty
Integrating both sides gives \begin {align*} \int \frac {1}{y^{3}}d y &= \int {dx}\\ -\frac {1}{2 y^{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 {1}{2}} = c_{1} -1 \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {1}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {1}{2}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {1}{2 y^{2}} = x +\frac {1}{2} \end {align*}
The constant \(c_{1} = {\frac {1}{2}}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} -\frac {1}{2 y^{2}} &= x +\frac {1}{2} \\
\end{align*} Verification of solutions
\[
-\frac {1}{2 y^{2}} = x +\frac {1}{2}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{3}=0, y \left (-1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{3}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 y^{2}}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {1}{\sqrt {-2 x -2 c_{1}}}, y=-\frac {1}{\sqrt {-2 x -2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=1 \\ {} & {} & 1=\frac {1}{\sqrt {2-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {1}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {1}{\sqrt {-1-2 x}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=1 \\ {} & {} & 1=-\frac {1}{\sqrt {2-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {1}{\sqrt {-1-2 x}} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 11
\[
y \left (x \right ) = \frac {1}{\sqrt {-2 x -1}}
\]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 14
\[
y(x)\to \frac {1}{\sqrt {-2 x-1}}
\]
8.11.2 Solving as quadrature ode
8.11.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful`
dsolve([diff(y(x),x)=y(x)^3,y(-1) = 1],y(x), singsol=all)
DSolve[{y'[x]==y[x]^3,{y[-1]==1}},y[x],x,IncludeSingularSolutions -> True]