Internal problem ID [14127]
Internal file name [OUTPUT/13808_Saturday_March_02_2024_02_50_50_PM_92682845/index.tex
]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.1, page 32
Problem number: 3.
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]
\[ \boxed {y^{\prime }-y^{\frac {1}{5}}=0} \] 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(t,y)\\ &= y^{\frac {1}{5}} \end {align*}
The \(y\) domain of \(f(t,y)\) when \(t=0\) 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 (y^{\frac {1}{5}}\right ) \\ &= \frac {1}{5 y^{\frac {4}{5}}} \end {align*}
The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) 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. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{\frac {1}{5}}=0, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{\frac {1}{5}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{\frac {1}{5}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{\frac {1}{5}}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {5 y^{\frac {4}{5}}}{4}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (\frac {4 c_{1}}{5}+\frac {4 t}{5}\right )^{\frac {5}{4}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\frac {4 \,4^{\frac {1}{4}} 5^{\frac {3}{4}} c_{1}^{\frac {5}{4}}}{25} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {4 \sqrt {2}\, 5^{\frac {3}{4}} t^{\frac {5}{4}}}{25} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {4 \sqrt {2}\, 5^{\frac {3}{4}} t^{\frac {5}{4}}}{25} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 5
\[
y \left (t \right ) = 0
\]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 24
\[
y(t)\to \frac {4 \sqrt {2} t^{5/4}}{5 \sqrt [4]{5}}
\]
3.4.2 Solving as quadrature ode
3.4.3 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(t),t)=y(t)^(1/5),y(0) = 0],y(t), singsol=all)
DSolve[{y'[t]==y[t]^(1/5),{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]