2.6 problem 6

Internal problem ID [4692]
Internal file name [OUTPUT/4185_Sunday_June_05_2022_12_38_28_PM_87658595/index.tex]

Book: A treatise on ordinary and partial differential equations by William Woolsey Johnson. 1913
Section: Chapter 2, Equations of the first order and degree. page 20
Problem number: 6.
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {y^{\prime }+b^{2} y^{2}=a^{2}} \]

2.6.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{-b^{2} y^{2}+a^{2}}d y &= x +c_{1}\\ -\frac {\ln \left (b y -a \right )-\ln \left (b y +a \right )}{2 b a}&=x +c_{1} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {a \left ({\mathrm e}^{-2 a b c_{1} -2 x b a}+1\right )}{b \left ({\mathrm e}^{-2 a b c_{1} -2 x b a}-1\right )}\\ &=-\frac {a \left (\frac {{\mathrm e}^{-2 x b a}}{c_{1}^{2}}+1\right )}{b \left (\frac {{\mathrm e}^{-2 x b a}}{c_{1}^{2}}-1\right )} \end {align*}

Summary

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

Verification of solutions

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

2.6.2 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 32

dsolve(diff(y(x),x)+b^2*y(x)^2=a^2,y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {a \left ({\mathrm e}^{-2 b a \left (x +c_{1} \right )}+1\right )}{b \left ({\mathrm e}^{-2 b a \left (x +c_{1} \right )}-1\right )} \]

✓ Solution by Mathematica

Time used: 3.208 (sec). Leaf size: 37

DSolve[y'[x]+b^2*y[x]^2==a^2,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {a \tanh (a b (x+c_1))}{b} \\ y(x)\to -\frac {a}{b} \\ y(x)\to \frac {a}{b} \\ \end{align*}