1.4 problem 4

1.4.1 Maple step by step solution

Internal problem ID [6770]
Internal file name [OUTPUT/6017_Monday_July_25_2022_01_59_37_AM_73862004/index.tex]

Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 94. Factoring the left member. EXERCISES Page 309
Problem number: 4.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "exact", "linear", "separable", "differentialType", "homogeneousTypeD2", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_separable]

\[ \boxed {x^{2} {y^{\prime }}^{2}+y^{\prime } x -y^{2}-y=0} \] The ode \begin {align*} x^{2} {y^{\prime }}^{2}+y^{\prime } x -y^{2}-y = 0 \end {align*}

is factored to \begin {align*} \left (y^{\prime } x +y+1\right ) \left (-y^{\prime } x +y\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} y^{\prime } x +y+1 = 0\tag {1} \\ -y^{\prime } x +y = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-y -1}{x} \end {align*}

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

Which simplifies to \begin {align*} \frac {1}{y +1} &= c_{2} x \end {align*}

Which simplifies to \[ y = -\frac {\left (c_{2} x \,{\mathrm e}^{c_{1}}-1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \]

Summary

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

Verification of solutions

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

Summary

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

Verification of solutions

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

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

Where \(f(x)=\frac {1}{x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x} \,d x}\\ \ln \left (y \right )&=\ln \left (x \right )+c_{3}\\ y&={\mathrm e}^{\ln \left (x \right )+c_{3}}\\ &=c_{3} x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x \\ \end{align*}

Verification of solutions

\[ y = c_{3} x \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x \\ \end{align*}

Verification of solutions

\[ y = c_{3} x \] Verified OK.

1.4.1 Maple step by step solution

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

Maple trace

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

Solution by Maple

Time used: 0.0 (sec). Leaf size: 19

dsolve(x^2*diff(y(x),x)^2+x*diff(y(x),x)-y(x)^2-y(x)=0,y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.045 (sec). Leaf size: 31

DSolve[x^2*(y'[x])^2+x*y'[x]-y[x]^2-y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

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