Internal problem ID [4752]
Internal file name [OUTPUT/4245_Sunday_June_05_2022_12_47_05_PM_37881317/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 2. Separable equations. page
398
Problem number: 4.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "bernoulli", "separable", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{2}+x y y^{\prime }=-1} \] With initial conditions \begin {align*} [y \left (5\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)\\ &= -\frac {y^{2}+1}{x y} \end {align*}
\(f(x,y)\) is not defined at \(y = 0\) therefore existence and uniqueness theorem do not apply.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= -\frac {y^{2}+1}{x y} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(y)=\frac {y^{2}+1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}+1}{y}} \,dy &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {y^{2}+1}{y}} \,dy} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (y^{2}+1\right )}{2}&=-\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {y^{2}+1} &= {\mathrm e}^{-\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {y^{2}+1} &= \frac {c_{2}}{x} \end {align*}
Which can be simplified to become \[ \sqrt {y^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{x} \] The solution is \[ \sqrt {y^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{x} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=5\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \frac {c_{2} {\mathrm e}^{c_{1}}}{5} \end {align*}
The solutions are \begin {align*} c_{1} = \ln \left (\frac {5}{c_{2}}\right ) \end {align*}
Trying the constant \begin {align*} c_{1} = \ln \left (\frac {5}{c_{2}}\right ) \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {y^{2}+1} = \frac {5}{x} \end {align*}
The constant \(c_{1} = \ln \left (\frac {5}{c_{2}}\right )\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {y^{2}+1} &= \frac {5}{x} \\ \end{align*}
Verification of solutions
\[ \sqrt {y^{2}+1} = \frac {5}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{2}+x y y^{\prime }=-1, y \left (5\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 }=\frac {-1-y^{2}}{x y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{-1-y^{2}}=\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{-1-y^{2}}d x =\int \frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (y^{2}+1\right )}{2}=\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {1-x^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} x}, y=-\frac {\sqrt {1-x^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} x}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (5\right )=0 \\ {} & {} & 0=\frac {\sqrt {1-25 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{5 \,{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\ln \left (5\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\ln \left (5\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {-x^{2}+25}}{x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (5\right )=0 \\ {} & {} & 0=-\frac {\sqrt {1-25 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{5 \,{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\ln \left (5\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\ln \left (5\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {-x^{2}+25}}{x} \\ \bullet & {} & \textrm {Solutions to the IVP}\hspace {3pt} \\ {} & {} & \left \{y=\frac {\sqrt {-x^{2}+25}}{x}, y=-\frac {\sqrt {-x^{2}+25}}{x}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 34
dsolve([(1+y(x)^2)+x*y(x)*diff(y(x),x)=0,y(5) = 0],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {-x^{2}+25}}{x} \\ y \left (x \right ) &= -\frac {\sqrt {-x^{2}+25}}{x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.329 (sec). Leaf size: 40
DSolve[{(1+y[x]^2)+x*y[x]*y'[x]==0,{y[5]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {25-x^2}}{x} \\ y(x)\to \frac {\sqrt {25-x^2}}{x} \\ \end{align*}