2.1.52 problem 52

Internal problem ID [8712]
Book : First order enumerated odes
Section : section 1
Problem number : 52
Date solved : Tuesday, December 17, 2024 at 12:58:16 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.

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}=\frac {y \left (x \right )^{2}}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {y \left (x \right )}{\sqrt {x}}, \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{\sqrt {x}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right )}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int \frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{2 \sqrt {x}+\textit {\_C1}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=-\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int -\frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=-2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{-2 \sqrt {x}+\textit {\_C1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )={\mathrm e}^{-2 \sqrt {x}+\mathit {C1}}, y \left (x \right )={\mathrm e}^{2 \sqrt {x}+\mathit {C1}}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying simple symmetries for implicit equations 
<- symmetries for implicit equations successful`
                                                                                    
                                                                                    
 

Maple dsolve solution

Solving time : 0.061 (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*}

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*}