Internal problem ID [2363]
Internal file name [OUTPUT/2363_Tuesday_February_27_2024_08_36_08_AM_76870095/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 39, page 179
Problem number: 32.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{2}-2 y^{\prime } x y+{y^{\prime }}^{2} \left (x^{2}-1\right )=0} \] The ode \begin {align*} y^{2}-2 y^{\prime } x y+{y^{\prime }}^{2} \left (x^{2}-1\right ) = 0 \end {align*}
is factored to \begin {align*} \left (-y^{\prime } x +y+y^{\prime }\right ) \left (-y^{\prime } x +y-y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} -y^{\prime } x +y+y^{\prime } = 0\tag {1} \\ -y^{\prime } x +y-y^{\prime } = 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}{x -1} \end {align*}
Where \(f(x)=\frac {1}{x -1}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x -1} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x -1} \,d x}\\ \ln \left (y \right )&=\ln \left (x -1\right )+c_{1}\\ y&={\mathrm e}^{\ln \left (x -1\right )+c_{1}}\\ &=c_{1} \left (x -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (x -1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (x -1\right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (x -1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (x -1\right ) \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y}{x +1} \end {align*}
Where \(f(x)=\frac {1}{x +1}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x +1} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x +1} \,d x}\\ \ln \left (y \right )&=\ln \left (x +1\right )+c_{2}\\ y&={\mathrm e}^{\ln \left (x +1\right )+c_{2}}\\ &=c_{2} \left (x +1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} \left (x +1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{2} \left (x +1\right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} \left (x +1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{2} \left (x +1\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2}-2 y^{\prime } x y+{y^{\prime }}^{2} \left (x^{2}-1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y}{x -1}, y^{\prime }=\frac {y}{x +1}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x -1} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x -1} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x -1}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x -1\right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} \left (x -1\right ) \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x +1} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x +1} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x +1}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x +1\right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} \left (x +1\right ) \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{c_{1}} \left (x -1\right ), y={\mathrm e}^{c_{1}} \left (x +1\right )\right \} \end {array} \]
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: 0.015 (sec). Leaf size: 17
dsolve(y(x)^2-2*diff(y(x),x)*x*y(x)+diff(y(x),x)^2*(x^2-1)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \left (x -1\right ) \\ y \left (x \right ) &= c_{1} \left (x +1\right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.106 (sec). Leaf size: 26
DSolve[y[x]^2-2*y'[x]*x*y[x]+y'[x]^2*(x^2-1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 (x-1) \\ y(x)\to c_1 (x+1) \\ y(x)\to 0 \\ \end{align*}