Internal problem ID [11367]
Internal file name [OUTPUT/10350_Wednesday_May_17_2023_07_49_44_PM_57195971/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.3.1 Separable equations. Exercises
page 26
Problem number: 1(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]
\[ \boxed {x^{\prime }-\sqrt {x}=0} \] With initial conditions \begin {align*} [x \left (0\right ) = 1] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} x^{\prime } &= f(t,x)\\ &= \sqrt {x} \end {align*}
The \(x\) domain of \(f(t,x)\) when \(t=0\) is \[ \{0\le x\} \] And the point \(x_0 = 1\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial x} &= \frac {\partial }{\partial x}\left (\sqrt {x}\right ) \\ &= \frac {1}{2 \sqrt {x}} \end {align*}
The \(x\) domain of \(\frac {\partial f}{\partial x}\) when \(t=0\) is \[
\{0
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {x}}d x &= \int {dt}\\ 2 \sqrt {x}&= t +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(x=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 2 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 2 \end {align*}
Trying the constant \begin {align*} c_{1} = 2 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} 2 \sqrt {x} = t +2 \end {align*}
The constant \(c_{1} = 2\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} 2 \sqrt {x} &= t +2 \\
\end{align*} Verification of solutions
\[
2 \sqrt {x} = t +2
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }-\sqrt {x}=0, x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=\sqrt {x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{\sqrt {x}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{\sqrt {x}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {x}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {1}{4} c_{1}^{2}+\frac {1}{2} c_{1} t +\frac {1}{4} t^{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=\frac {c_{1}^{2}}{4} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (-2, 2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\left (-2, 2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {\left (t -2\right )^{2}}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {\left (t -2\right )^{2}}{4} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 11
\[
x \left (t \right ) = \frac {\left (t +2\right )^{2}}{4}
\]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 16
\[
x(t)\to \frac {2 t^{3/2}}{3}+1
\]
4.1.2 Solving as quadrature ode
4.1.3 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(x(t),t)=sqrt(x(t)),x(0) = 1],x(t), singsol=all)
DSolve[{x'[t]==Sqrt[t],{x[0]==1}},x[t],t,IncludeSingularSolutions -> True]