Internal problem ID [12272]
Internal file name [OUTPUT/10925_Thursday_September_28_2023_01_09_06_AM_20167140/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Differential Equations. Problems page
221
Problem number: Problem 19(e).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (1-y\right ) y^{\prime \prime }-{y^{\prime }}^{2}=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (1-y\right ) y^{\prime \prime }-{y^{\prime }}^{2}\right )d x &= 0 \\ -\left (-1+y\right ) y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int -\frac {-1+y}{c_{1}}d y &= x +c_{2}\\ -\frac {y \left (y -2\right )}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1}\\ y_2&=1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \\ \tag{2} y &= 1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \\ \end{align*}
Verification of solutions
\[ y = 1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \] Verified OK.
\[ y = 1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \] Verified OK.
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 (1-y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-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 {p}{-1+y} \end {align*}
Where \(f(y)=-\frac {1}{-1+y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {1}{-1+y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {1}{-1+y} \,d y}\\ \ln \left (p \right )&=-\ln \left (-1+y \right )+c_{1}\\ p&={\mathrm e}^{-\ln \left (-1+y \right )+c_{1}}\\ &=\frac {c_{1}}{-1+y} \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 } = \frac {c_{1}}{-1+y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {-1+y}{c_{1}}d y &= x +c_{2}\\ \frac {y \left (y -2\right )}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=1-\sqrt {2 c_{1} c_{2} +2 c_{1} x +1}\\ y_2&=1+\sqrt {2 c_{1} c_{2} +2 c_{1} x +1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1-\sqrt {2 c_{1} c_{2} +2 c_{1} x +1} \\ \tag{2} y &= 1+\sqrt {2 c_{1} c_{2} +2 c_{1} x +1} \\ \end{align*}
Verification of solutions
\[ y = 1-\sqrt {2 c_{1} c_{2} +2 c_{1} x +1} \] Verified OK.
\[ y = 1+\sqrt {2 c_{1} c_{2} +2 c_{1} x +1} \] Verified OK.
Writing the ode as \[ \left (1-y\right ) y^{\prime \prime }-{y^{\prime }}^{2} = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (1-y\right ) y^{\prime \prime }-{y^{\prime }}^{2}\right )d x &= 0 \\ -y y^{\prime }+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int -\frac {-1+y}{c_{1}}d y &= x +c_{2}\\ -\frac {y \left (y -2\right )}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1}\\ y_2&=1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \\ \tag{2} y &= 1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \\ \end{align*}
Verification of solutions
\[ y = 1-\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \] Verified OK.
\[ y = 1+\sqrt {-2 c_{1} c_{2} -2 c_{1} x +1} \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= 1-y\\ a_1 &= -y^{\prime }\\ a_0 &= 0 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1-y\,d y'} + \int {-y^{\prime }\,d y} + \int {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} -2 y y^{\prime }+y^{\prime } = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int -\frac {2 y -1}{c_{1}}d y &= x +c_{2}\\ -\frac {y \left (-1+y \right )}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {1}{2}-\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2}\\ y_2&=\frac {1}{2}+\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{2}-\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2} \\ \tag{2} y &= \frac {1}{2}+\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{2}-\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2} \] Verified OK.
\[ y = \frac {1}{2}+\frac {\sqrt {-4 c_{1} c_{2} -4 c_{1} x +1}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (1-y\right ) \left (\frac {d}{d x}y^{\prime }\right )-{y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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 (1-y \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )}{1-y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {1}{1-y} \\ \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 {1}{1-y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=-\ln \left (1-y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {{\mathrm e}^{c_{1}}}{-1+y} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {{\mathrm e}^{c_{1}}}{-1+y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{c_{1}}}{-1+y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{c_{1}}}{-1+y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (-1+y\right ) y^{\prime }=-{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (-1+y\right ) y^{\prime }d x =\int -{\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}-y=-{\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=1-\sqrt {1-2 \,{\mathrm e}^{c_{1}} x +2 c_{2}}, y=1+\sqrt {1-2 \,{\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: 0.031 (sec). Leaf size: 39
dsolve((1-y(x))*diff(y(x),x$2)-diff(y(x),x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 1 \\ y \left (x \right ) &= 1-\sqrt {2 c_{1} x +2 c_{2} +1} \\ y \left (x \right ) &= 1+\sqrt {2 c_{1} x +2 c_{2} +1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.881 (sec). Leaf size: 49
DSolve[(1-y[x])*y''[x]- y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 1-\sqrt {-2 c_1 x+1-2 c_2 c_1} \\ y(x)\to 1+\sqrt {-2 c_1 x+1-2 c_2 c_1} \\ \end{align*}