Internal problem ID [4656]
Internal file name [OUTPUT/4149_Sunday_June_05_2022_12_29_20_PM_35294735/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.6, page 504.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (1+y\right ) y^{\prime \prime }-3 {y^{\prime }}^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} \left (y +1\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-3 p \left (y \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {3 p}{y +1} \end {align*}
Where \(f(y)=\frac {3}{y +1}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {3}{y +1} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {3}{y +1} \,d y}\\ \ln \left (p \right )&=3 \ln \left (y +1\right )+c_{1}\\ p&={\mathrm e}^{3 \ln \left (y +1\right )+c_{1}}\\ &=c_{1} \left (y +1\right )^{3} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = c_{1} \left (1+y\right )^{3} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{1} \left (y +1\right )^{3}}d y &= x +c_{2}\\ -\frac {1}{2 \left (y +1\right )^{2} c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}-1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}}\\ y_2&=-\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}+1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}-1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}} \\ \tag{2} y &= -\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}+1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}-1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}} \] Verified OK.
\[ y = -\frac {\sqrt {-2 c_{1} c_{2} -2 c_{1} x}+1}{\sqrt {-2 c_{1} c_{2} -2 c_{1} x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (1+y\right ) y^{\prime \prime }-3 {y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (y +1\right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-3 u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {3 u \left (y \right )}{y +1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {3}{y +1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {3}{y +1}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=3 \ln \left (y +1\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (y +1\right )^{3} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (y +1\right )^{3} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (1+y\right )^{3} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (1+y\right )^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\left (1+y\right )^{3}}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\left (1+y\right )^{3}}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 \left (1+y\right )^{2}}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}-1}{\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}}, y=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}+1}{\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 59
dsolve((y(x)+1)*diff(y(x),x$2)=3*(diff(y(x),x))^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -1 \\ y \left (x \right ) &= -\frac {\sqrt {-2 c_{1} x -2 c_{2}}-1}{\sqrt {-2 c_{1} x -2 c_{2}}} \\ y \left (x \right ) &= -\frac {\sqrt {-2 c_{1} x -2 c_{2}}+1}{\sqrt {-2 c_{1} x -2 c_{2}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.485 (sec). Leaf size: 107
DSolve[(y[x]+1)*y''[x]==3*(y'[x])^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {2 c_1 x+\sqrt {2} \sqrt {-c_1 (x+c_2)}+2 c_2 c_1}{2 c_1 (x+c_2)} \\ y(x)\to \frac {-2 c_1 x+\sqrt {2} \sqrt {-c_1 (x+c_2)}-2 c_2 c_1}{2 c_1 (x+c_2)} \\ y(x)\to -1 \\ y(x)\to \text {Indeterminate} \\ \end{align*}