26.13 problem 749

26.13.1 Maple step by step solution

Internal problem ID [3990]
Internal file name [OUTPUT/3483_Sunday_June_05_2022_09_24_57_AM_65175101/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 26
Problem number: 749.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "quadrature"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {{y^{\prime }}^{2}-y=0} \] Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {y} \tag {1} \\ y^{\prime }&=-\sqrt {y} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {y}}d y &= \int d x \\ 2 \sqrt {y}&=x +c_{1} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} 2 \sqrt {y} &= x +c_{1} \\ \end{align*}

Verification of solutions

\[ 2 \sqrt {y} = x +c_{1} \] Verified OK.

Solving equation (2)

Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {y}}d y &= \int d x \\ -2 \sqrt {y}&=x +c_{2} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} -2 \sqrt {y} &= x +c_{2} \\ \end{align*}

Verification of solutions

\[ -2 \sqrt {y} = x +c_{2} \] Verified OK.

26.13.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\sqrt {y}, y^{\prime }=-\sqrt {y}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\sqrt {y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y}}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {y}=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {1}{4} x^{2}+\frac {1}{2} c_{1} x +\frac {1}{4} c_{1}^{2} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\sqrt {y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y}}d x =\int \left (-1\right )d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {y}=c_{1} -x \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {1}{4} x^{2}-\frac {1}{2} c_{1} x +\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {1}{4} x^{2}-\frac {1}{2} c_{1} x +\frac {1}{4} c_{1}^{2}, y=\frac {1}{4} x^{2}+\frac {1}{2} c_{1} x +\frac {1}{4} c_{1}^{2}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
   *** Sublevel 2 *** 
   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 
   <- differential order: 1; missing  x  successful`
 

Solution by Maple

Time used: 0.078 (sec). Leaf size: 17

dsolve(diff(y(x),x)^2 = y(x),y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {\left (-c_{1} +x \right )^{2}}{4} \\ \end{align*}

Solution by Mathematica

Time used: 0.061 (sec). Leaf size: 36

DSolve[(y'[x])^2 == y[x],y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {1}{4} (x-c_1){}^2 \\ y(x)\to \frac {1}{4} (x+c_1){}^2 \\ y(x)\to 0 \\ \end{align*}