Internal problem ID [13297]
Internal file name [OUTPUT/12469_Wednesday_February_14_2024_02_06_31_AM_67567237/index.tex
]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 3. Some basics about First order equations. Additional exercises. page
63
Problem number: 3.6.
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]
\[ \boxed {y^{\prime }-2 \sqrt {y}=0} \] With initial conditions \begin {align*} [y \left (1\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)\\ &= 2 \sqrt {y} \end {align*}
The \(y\) domain of \(f(x,y)\) when \(x=1\) is \[ \{0\le y\} \] And the point \(y_0 = 0\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (2 \sqrt {y}\right ) \\ &= \frac {1}{\sqrt {y}} \end {align*}
The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=1\) is \[
\{0
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 0\) is verified to satisfy the ode, then the solution is
\begin {align*} y&=y_0 \\ &=0 \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= 0 \\
\end{align*} Verification of solutions
\[
y = 0
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 \sqrt {y}=0, y \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 \sqrt {y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y}}=2 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y}}d x =\int 2d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {y}=2 x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x^{2}+c_{1} x +\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=1+c_{1} +\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (-2, -2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\left (-2, -2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (x -1\right )^{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (x -1\right )^{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 5
\[
y \left (x \right ) = 0
\]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 6
\[
y(x)\to 0
\]
2.11.2 Solving as quadrature ode
2.11.3 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)=2*sqrt(y(x)),y(1) = 0],y(x), singsol=all)
DSolve[{y'[x]==2*Sqrt[y[x]],{y[1]==0}},y[x],x,IncludeSingularSolutions -> True]