2.7 problem 7

Internal problem ID [4755]
Internal file name [OUTPUT/4248_Sunday_June_05_2022_12_47_39_PM_22574787/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: 7.
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]

\[ \boxed {y y^{\prime }+x y^{2}=8 x} \] With initial conditions \begin {align*} [y \left (1\right ) = 3] \end {align*}

2.7.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\frac {x \left (y^{2}-8\right )}{y} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=3\) is \[ \{-\infty

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=3\) is \[ \{-\infty

2.7.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= -\frac {x \left (y^{2}-8\right )}{y} \end {align*}

Where \(f(x)=-x\) and \(g(y)=\frac {y^{2}-8}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}-8}{y}} \,dy &= -x \,d x \\ \int { \frac {1}{\frac {y^{2}-8}{y}} \,dy} &= \int {-x \,d x} \\ \frac {\ln \left (y^{2}-8\right )}{2}&=-\frac {x^{2}}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {y^{2}-8} &= {\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \end {align*}

Which simplifies to \begin {align*} \sqrt {y^{2}-8} &= c_{2} {\mathrm e}^{-\frac {x^{2}}{2}} \end {align*}

The solution is \[ \sqrt {y^{2}-8} = c_{2} {\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = c_{2} {\mathrm e}^{-\frac {1}{2}+c_{1}} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {1}{2}-\ln \left (c_{2} \right ) \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {1}{2}-\ln \left (c_{2} \right ) \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {y^{2}-8} = {\mathrm e}^{-\frac {\left (x -1\right ) \left (x +1\right )}{2}} \end {align*}

The constant \(c_{1} = \frac {1}{2}-\ln \left (c_{2} \right )\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \sqrt {y^{2}-8} &= {\mathrm e}^{-\frac {\left (x -1\right ) \left (x +1\right )}{2}} \\ \end{align*}

Verification of solutions

\[ \sqrt {y^{2}-8} = {\mathrm e}^{-\frac {\left (x -1\right ) \left (x +1\right )}{2}} \] Verified OK.

2.7.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime }+x y^{2}=8 x , y \left (1\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-x y^{2}+8 x}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{y^{2}-8}=-x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{y^{2}-8}d x =\int -x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y^{2}-8\right )}{2}=-\frac {x^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {8+{\mathrm e}^{-x^{2}+2 c_{1}}}, y=-\sqrt {8+{\mathrm e}^{-x^{2}+2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\sqrt {8+{\mathrm e}^{-1+2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {1}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {8+{\mathrm e}^{-\left (x -1\right ) \left (x +1\right )}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=-\sqrt {8+{\mathrm e}^{-1+2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\sqrt {8+{\mathrm e}^{-\left (x -1\right ) \left (x +1\right )}} \end {array} \]

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: 0.063 (sec). Leaf size: 17

dsolve([y(x)*diff(y(x),x)+(x*y(x)^2-8*x)=0,y(1) = 3],y(x), singsol=all)
 

\[ y \left (x \right ) = \sqrt {{\mathrm e}^{-\left (x -1\right ) \left (1+x \right )}+8} \]

✓ Solution by Mathematica

Time used: 1.924 (sec). Leaf size: 39

DSolve[{y[x]*y'[x]+(x*y[x]^2-8*x)==0,{y[1]==3}},y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \sqrt {e^{1-x^2}+8} \\ y(x)\to \sqrt {e^{1-x^2}+8} \\ \end{align*}