Internal problem ID [15129]
Internal file name [OUTPUT/15130_Sunday_April_21_2024_01_36_45_PM_99181376/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Section 11. Singular solutions of differential equations. Exercises page 92
Problem number: 268.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (y^{\prime }-1\right )^{2}-y^{2}=0} \] The ode \begin {align*} \left (y^{\prime }-1\right )^{2}-y^{2} = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime }-1+y\right ) \left (y^{\prime }-1-y\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime }-1+y = 0\tag {1} \\ y^{\prime }-1-y = 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 +1}d y &= x +c_{1}\\ -\ln \left (y -1\right )&=x +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{-x -c_{1}}+1\\ &=\frac {{\mathrm e}^{-x}}{c_{1}}+1 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{1}}+1 \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{1}}+1 \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{1}}+1 \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{1}}+1 \] Verified OK.
Solving ODE (2) Integrating both sides gives \begin {align*} \int \frac {1}{y +1}d y &= x +c_{2}\\ \ln \left (y +1\right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{x +c_{2}}-1\\ &={\mathrm e}^{x} c_{2} -1 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} c_{2} -1 \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} c_{2} -1 \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} c_{2} -1 \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} c_{2} -1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y^{\prime }-1\right )^{2}-y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-y+1, y^{\prime }=y+1\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-y+1 \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-y+1}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-y+1}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (-y+1\right )=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-{\mathrm e}^{-x -c_{1}}+1 \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=y+1 \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y+1}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y+1}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y+1\right )=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{x +c_{1}}-1 \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=-{\mathrm e}^{-x -c_{1}}+1, y={\mathrm e}^{x +c_{1}}-1\right \} \end {array} \]
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: 0.016 (sec). Leaf size: 21
dsolve((diff(y(x),x)-1)^2=y(x)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -1+{\mathrm e}^{x} c_{1} \\ y \left (x \right ) &= 1+c_{1} {\mathrm e}^{-x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.074 (sec). Leaf size: 37
DSolve[(y'[x]-1)^2==y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 1+c_1 e^{-x} \\ y(x)\to -1+c_1 e^x \\ y(x)\to -1 \\ y(x)\to 1 \\ \end{align*}