1.52 problem 52
Internal
problem
ID
[8016]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
52
Date
solved
:
Monday, October 21, 2024 at 04:40:57 PM
CAS
classification
:
[_separable]
Solve
\begin{align*} {y^{\prime }}^{2}&=\frac {y^{2}}{x} \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {y}{\sqrt {x}} \\
\tag{2} y^{\prime }&=-\frac {y}{\sqrt {x}} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=-\frac {1}{\sqrt {x}}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {1}{\sqrt {x}}d x}\\ &= {\mathrm e}^{-2 \sqrt {x}} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{-2 \sqrt {x}}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} y \,{\mathrm e}^{-2 \sqrt {x}}&= \int {0 \,dx} + c_2 \\ &=c_2 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{-2 \sqrt {x}}\) gives the final solution
\[ y = {\mathrm e}^{2 \sqrt {x}} c_2 \]
We now need to find
the singular solutions, these are found by finding for what values \((\frac {y}{\sqrt {x}})\) is zero. These give
\begin{align*}
y&=0 \\
\end{align*}
Now
we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(y = 0\) satisfies the ode and initial conditions.
Solving Eq. (2)
In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{\sqrt {x}}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{\sqrt {x}}d x}\\ &= {\mathrm e}^{2 \sqrt {x}} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{2 \sqrt {x}}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} y \,{\mathrm e}^{2 \sqrt {x}}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{2 \sqrt {x}}\) gives the final solution
\[ y = {\mathrm e}^{-2 \sqrt {x}} c_3 \]
We now need to find
the singular solutions, these are found by finding for what values \((-\frac {y}{\sqrt {x}})\) is zero. These give
\begin{align*}
y&=0 \\
\end{align*}
Now
we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(y = 0\) satisfies the ode and initial conditions.
1.52.1 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}=\frac {y^{2}}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y}{\sqrt {x}}, y^{\prime }=-\frac {y}{\sqrt {x}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{2 \sqrt {x}+\textit {\_C1}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int -\frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-2 \sqrt {x}+\textit {\_C1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{-2 \sqrt {x}+\mathit {C1}}, y={\mathrm e}^{2 \sqrt {x}+\mathit {C1}}\right \} \end {array} \]
1.52.2 Maple trace
Methods for first order ODEs:
1.52.3 Maple dsolve solution
Solving time : 0.027 (sec)
Leaf size : 27
dsolve(diff(y(x),x)^2 = y(x)^2/x,
y(x),singsol=all)
\begin{align*}
y &= 0 \\
y &= c_1 \,{\mathrm e}^{-2 \sqrt {x}} \\
y &= c_1 \,{\mathrm e}^{2 \sqrt {x}} \\
\end{align*}
1.52.4 Mathematica DSolve solution
Solving time : 0.067 (sec)
Leaf size : 38
DSolve[{(D[y[x],x])^2==y[x]^2/x,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to c_1 e^{-2 \sqrt {x}} \\
y(x)\to c_1 e^{2 \sqrt {x}} \\
y(x)\to 0 \\
\end{align*}