Internal problem ID [5356]
Internal file name [OUTPUT/4847_Sunday_February_04_2024_12_46_27_AM_79210250/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 12. Linear equations of order n. Supplemetary problems. Page 81
Problem number: 18.
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], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{2}=2} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= \int 2d x\\ y y^{\prime } = 2 x + c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{1} +2 x}{y} \end {align*}
Where \(f(x)=c_{1} +2 x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= c_{1} +2 x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {c_{1} +2 x \,d x} \\ \frac {y^{2}}{2}&=c_{1} x +x^{2}+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} = 0 \] 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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = 2 \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^{2}-2}{y p} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {p^{2}-2}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}-2}{p}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}-2}{p}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}-2\right )}{2}&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}-2} &= {\mathrm e}^{-\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}-2} &= \frac {c_{2}}{y} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}-2} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] The solution is \[ \sqrt {p \left (y \right )^{2}-2} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}-2} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}d y &= \int d x \\ \frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}d y &= \int d x \\ -\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2}&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2} &= x +c_{3} \\ \tag{2} -\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2} = x +c_{3} \] Verified OK.
\[ -\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 y^{2}}}{2} = x +c_{4} \] Verified OK.
Writing the ode as \[ y y^{\prime \prime }+{y^{\prime }}^{2} = 2 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= \int 2d x\\ y y^{\prime } = 2 x +c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{1} +2 x}{y} \end {align*}
Where \(f(x)=c_{1} +2 x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= c_{1} +2 x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {c_{1} +2 x \,d x} \\ \frac {y^{2}}{2}&=c_{1} x +x^{2}+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-c_{1} x -x^{2}-c_{2} = 0 \] 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 &= y\\ a_1 &= y^{\prime }\\ a_0 &= -2 \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 {y\,d y'} + \int {y^{\prime }\,d y} + \int {-2\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} 2 y y^{\prime }-2 x = c_{1} \end {align*}
Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\frac {c_{1}}{2}+x}{y} \end {align*}
Where \(f(x)=\frac {c_{1}}{2}+x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= \frac {c_{1}}{2}+x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {\frac {c_{1}}{2}+x \,d x} \\ \frac {y^{2}}{2}&=\frac {1}{2} c_{1} x +\frac {1}{2} x^{2}+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-\frac {c_{1} x}{2}-\frac {x^{2}}{2}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-\frac {c_{1} x}{2}-\frac {x^{2}}{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-\frac {c_{1} x}{2}-\frac {x^{2}}{2}-c_{2} = 0 \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime \prime }+{y^{\prime }}^{2}=2 \\ \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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+u \left (y \right )^{2}=2 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {-u \left (y \right )^{2}+2}{y u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}+2}=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}+2}d y =\int \frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (y \right )^{2}-2\right )}{2}=\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y}, u \left (y \right )=-\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}=\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}d x =\int \frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{2 \left ({\mathrm e}^{c_{1}}\right )^{2}}=\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2+8 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+16 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +8 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}, y=\frac {\sqrt {-2+8 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+16 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +8 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}=-\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}d x =\int -\frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+1}}{2 \left ({\mathrm e}^{c_{1}}\right )^{2}}=-\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2+8 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-16 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +8 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}, y=\frac {\sqrt {-2+8 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-16 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +8 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature <- quadrature successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 39
dsolve(y(x)*diff(y(x),x$2)+diff(y(x),x)^2=2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {-2 c_{1} x +2 x^{2}+2 c_{2}} \\ y \left (x \right ) &= -\sqrt {-2 c_{1} x +2 x^{2}+2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 6.295 (sec). Leaf size: 101
DSolve[y[x]*y''[x]+y'[x]^2==2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {4 (x+c_2){}^2-e^{2 c_1}}}{\sqrt {2}} \\ y(x)\to \sqrt {2 (x+c_2){}^2-\frac {e^{2 c_1}}{2}} \\ y(x)\to -\sqrt {2} \sqrt {(x+c_2){}^2} \\ y(x)\to \sqrt {2} \sqrt {(x+c_2){}^2} \\ \end{align*}