18.24 problem 24

Internal problem ID [2296]
Internal file name [OUTPUT/2296_Tuesday_February_27_2024_08_24_13_AM_1427113/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 35, page 157
Problem number: 24.
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", "second_order_nonlinear_solved_by_mainardi_lioville_method"

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]]

\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{2}-y y^{\prime }=0} \]

18.24.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+\left (y^{\prime }-y\right ) y^{\prime }\right )d x &= 0 \\ -\frac {y^{2}}{2}+y y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {2 y}{y^{2}+2 c_{1}}d y &= x +c_{2}\\ \ln \left (y^{2}+2 c_{1} \right )&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}}\\ y_2&=-\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=-\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \tag{2} y &= -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \end{align*}

Verification of solutions

\[ y = \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

\[ y = -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

18.24.2 Solving as second order ode missing x ode

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 )+\left (p \left (y \right )-y \right ) p \left (y \right ) = 0 \end {align*}

Which is now solved as first order ode for \(p(y)\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d y}p \left (y \right ) + p(y)p \left (y \right ) &= q(y) \end {align*}

Where here \begin {align*} p(y) &=\frac {1}{y}\\ q(y) &=1 \end {align*}

Hence the ode is \begin {align*} \frac {d}{d y}p \left (y \right )+\frac {p \left (y \right )}{y} = 1 \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {1}{y}d y} \\ &= y \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu p\right ) &= \mu \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (y p\right ) &= y\\ \mathrm {d} \left (y p\right ) &= y\mathrm {d} y \end {align*}

Integrating gives \begin {align*} y p &= \int {y\,\mathrm {d} y}\\ y p &= \frac {y^{2}}{2} + c_{1} \end {align*}

Dividing both sides by the integrating factor \(\mu =y\) results in \begin {align*} p \left (y \right ) &= \frac {y}{2}+\frac {c_{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 {y}{2}+\frac {c_{1}}{y} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {2 y}{y^{2}+2 c_{1}}d y &= x +c_{2}\\ \ln \left (y^{2}+2 c_{1} \right )&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}}\\ y_2&=-\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=-\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \tag{2} y &= -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \end{align*}

Verification of solutions

\[ y = \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

\[ y = -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

18.24.3 Solving as second order nonlinear solved by mainardi lioville method ode

The ode has the Liouville form given by \begin {align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end {align*}

Where in this problem \begin {align*} f(x) &= -1\\ g(y) &= \frac {1}{y} \end {align*}

Dividing through by \(y^{\prime }\) then Eq (1A) becomes \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end {align*}

But the first term in Eq (2A) can be written as \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end {align*}

And the last term in Eq (2A) can be written as \begin {align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end {align*}

Substituting (3A,4A) back into (2A) gives \begin {align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end {align*}

Integrating the above w.r.t. \(x\) gives \begin {align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_{1} \end {align*}

Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives \begin {align*} y^{\prime } &= c_{2} e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end {align*}

Where \(c_{2}\) is a new arbitrary constant. But since \(g=\frac {1}{y}\) and \(f=-1\), then \begin {align*} \int -g d y &= \int -\frac {1}{y}d y\\ &= -\ln \left (y\right )\\ \int -f d x &= \int 1d x\\ &= x \end {align*}

Substituting the above into Eq(6A) gives \[ y^{\prime } = \frac {c_{2} {\mathrm e}^{x}}{y} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{2} {\mathrm e}^{x}}{y} \end {align*}

Where \(f(x)=c_{2} {\mathrm e}^{x}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= c_{2} {\mathrm e}^{x} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {c_{2} {\mathrm e}^{x} \,d x} \\ \frac {y^{2}}{2}&=c_{2} {\mathrm e}^{x}+c_{3} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{2} {\mathrm e}^{x}-c_{3} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-c_{2} {\mathrm e}^{x}-c_{3} &= 0 \\ \end{align*}

Verification of solutions

\[ \frac {y^{2}}{2}-c_{2} {\mathrm e}^{x}-c_{3} = 0 \] Verified OK.

18.24.4 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y y^{\prime \prime }+\left (y^{\prime }-y\right ) y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+\left (y^{\prime }-y\right ) y^{\prime }\right )d x &= 0 \\ -\frac {y^{2}}{2}+y y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {2 y}{y^{2}+2 c_{1}}d y &= x +c_{2}\\ \ln \left (y^{2}+2 c_{1} \right )&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}}\\ y_2&=-\sqrt {-2 c_{1} +{\mathrm e}^{x +c_{2}}}\\ &=-\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \tag{2} y &= -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \\ \end{align*}

Verification of solutions

\[ y = \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

\[ y = -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{x}} \] Verified OK.

18.24.5 Solving as exact nonlinear second order ode ode

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 }-y\\ 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 {y\,d y'} + \int {y^{\prime }-y\,d y} + \int {0\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} 2 y y^{\prime }-\frac {y^{2}}{2} = c_{1} \end {align*}

Which is now solved Integrating both sides gives \begin {align*} \int \frac {4 y}{y^{2}+2 c_{1}}d y &= x +c_{2}\\ 2 \ln \left (y^{2}+2 c_{1} \right )&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {-2 c_{1} +{\mathrm e}^{\frac {x}{2}+\frac {c_{2}}{2}}}\\ &=\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}}\\ y_2&=-\sqrt {-2 c_{1} +{\mathrm e}^{\frac {x}{2}+\frac {c_{2}}{2}}}\\ &=-\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}} \\ \tag{2} y &= -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}} \\ \end{align*}

Verification of solutions

\[ y = \sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}} \] Verified OK.

\[ y = -\sqrt {-2 c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}}} \] Verified OK.

18.24.6 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime \prime }+\left (y^{\prime }-y\right ) y^{\prime }=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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\left (u \left (y \right )-y \right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {u \left (y \right )-y}{y} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=1-\frac {u \left (y \right )}{y} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )+\frac {u \left (y \right )}{y}=1 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (y \right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )+\frac {u \left (y \right )}{y}\right )=\mu \left (y \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )+\frac {u \left (y \right )}{y}\right )=\left (\frac {d}{d y}u \left (y \right )\right ) \mu \left (y \right )+u \left (y \right ) \left (\frac {d}{d y}\mu \left (y \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d y}\mu \left (y \right ) \\ {} & {} & \frac {d}{d y}\mu \left (y \right )=\frac {\mu \left (y \right )}{y} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (y \right )=y \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right )\right )d y =\int \mu \left (y \right )d y +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (y \right ) \mu \left (y \right )=\int \mu \left (y \right )d y +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\int \mu \left (y \right )d y +c_{1}}{\mu \left (y \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (y \right )=y \\ {} & {} & u \left (y \right )=\frac {\int y d y +c_{1}}{y} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & u \left (y \right )=\frac {\frac {y^{2}}{2}+c_{1}}{y} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (y \right )=\frac {y^{2}+2 c_{1}}{2 y} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y^{2}+2 c_{1}}{2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {y^{2}+2 c_{1}}{2 y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+2 c_{1}}{2 y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{y^{2}+2 c_{1}}=\frac {1}{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{y^{2}+2 c_{1}}d x =\int \frac {1}{2}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y^{2}+2 c_{1} \right )}{2}=\frac {x}{2}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-2 c_{1} +{\mathrm e}^{x +2 c_{2}}}, y=-\sqrt {-2 c_{1} +{\mathrm e}^{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: 0.047 (sec). Leaf size: 35

dsolve(y(x)*diff(y(x),x$2)+diff(y(x),x)^2=y(x)*diff(y(x),x),y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \sqrt {2 \,{\mathrm e}^{x} c_{1} +2 c_{2}} \\ y \left (x \right ) &= -\sqrt {2 \,{\mathrm e}^{x} c_{1} +2 c_{2}} \\ \end{align*}

✓ Solution by Mathematica

Time used: 1.093 (sec). Leaf size: 41

DSolve[y[x]*y''[x]+y'[x]^2==y[x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to c_2 \sqrt {2 e^x+e^{c_1}} \\ y(x)\to \sqrt {2} c_2 \sqrt {e^x} \\ \end{align*}