1.32 problem 33

Internal problem ID [7076]
Internal file name [OUTPUT/6062_Sunday_June_05_2022_04_16_50_PM_96181182/index.tex]

Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 33.
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]

\[ \boxed {y^{\prime }-\frac {1}{1-y}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2] \end {align*}

1.32.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= -\frac {1}{-1+y} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{y <1\boldsymbol {\lor }1

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{y <1\boldsymbol {\lor }1

1.32.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \left (1-y \right )d y &= \int {dt}\\ -\frac {y \left (-2+y \right )}{2}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=2\) 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*} -\frac {y \left (-2+y \right )}{2} = t \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {y \left (-2+y\right )}{2} &= t \\ \end{align*}

Verification of solutions

\[ -\frac {y \left (-2+y\right )}{2} = t \] Verified OK.

1.32.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {1}{1-y}=0, y \left (0\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{1-y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (1-y\right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime } \left (1-y\right )d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {y^{2}}{2}+y=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=1-\sqrt {1-2 c_{1} -2 t}, y=1+\sqrt {1-2 c_{1} -2 t}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=1-\sqrt {1-2 c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=1+\sqrt {1-2 c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=1+\sqrt {1-2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=1+\sqrt {1-2 t} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 13

dsolve([diff(y(t),t)=1/(1-y(t)),y(0) = 2],y(t), singsol=all)
 

\[ y \left (t \right ) = 1+\sqrt {1-2 t} \]

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 16

DSolve[{y'[t]==1/(1-y[t]),y[0]==2},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \sqrt {1-2 t}+1 \]