Internal problem ID [6780]
Internal file name [OUTPUT/6027_Monday_July_25_2022_01_59_58_AM_88349000/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: 14.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "quadrature", "separable", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {y {y^{\prime }}^{2} x +\left (x y^{2}-1\right ) y^{\prime }-y=0} \] The ode \begin {align*} y {y^{\prime }}^{2} x +\left (x y^{2}-1\right ) y^{\prime }-y = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime }+y\right ) \left (y^{\prime } x y-1\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime }+y = 0\tag {1} \\ y^{\prime } x y-1 = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} \int -\frac {1}{y}d y &= x +c_{1}\\ -\ln \left (y \right )&=x +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{-x -c_{1}}\\ &=\frac {{\mathrm e}^{-x}}{c_{1}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{1}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{1}} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {1}{y x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {\frac {1}{x} \,d x} \\ \frac {y^{2}}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {2 \ln \left (x \right )+2 c_{2}} \\ y &= -\sqrt {2 \ln \left (x \right )+2 c_{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 \ln \left (x \right )+2 c_{2}} \\ \tag{2} y &= -\sqrt {2 \ln \left (x \right )+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 \ln \left (x \right )+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {2 \ln \left (x \right )+2 c_{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 \ln \left (x \right )+2 c_{2}} \\ \tag{2} y &= -\sqrt {2 \ln \left (x \right )+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 \ln \left (x \right )+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {2 \ln \left (x \right )+2 c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y {y^{\prime }}^{2} x +\left (x y^{2}-1\right ) y^{\prime }-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-y, y^{\prime }=\frac {1}{y x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-y \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \left (-1\right )d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-x +c_{1}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {1}{y x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {2 \ln \left (x \right )+2 c_{1}}, y=-\sqrt {2 \ln \left (x \right )+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{-x +c_{1}}, \left \{y=\sqrt {2 \ln \left (x \right )+2 c_{1}}, y=-\sqrt {2 \ln \left (x \right )+2 c_{1}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli 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: 0.016 (sec). Leaf size: 34
dsolve(x*y(x)*diff(y(x),x)^2+(x*y(x)^2-1)*diff(y(x),x)-y(x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {2 \ln \left (x \right )+c_{1}} \\ y \left (x \right ) &= -\sqrt {2 \ln \left (x \right )+c_{1}} \\ y \left (x \right ) &= c_{1} {\mathrm e}^{-x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.108 (sec). Leaf size: 57
DSolve[x*y[x]*(y'[x])^2+(x*y[x]^2-1)*y'[x]-y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 e^{-x} \\ y(x)\to -\sqrt {2} \sqrt {\log (x)+c_1} \\ y(x)\to \sqrt {2} \sqrt {\log (x)+c_1} \\ y(x)\to 0 \\ \end{align*}