2.2.10 Problem 11

Internal problem ID [10421]
Book : Second order enumerated odes
Section : section 2
Problem number : 11
Date solved : Monday, December 08, 2025 at 08:50:10 PM
CAS classification : [[_2nd_order, _missing_y], _Liouville, [_2nd_order, _reducible, _mu_xy]]

2.2.10.1 second order ode missing y

0.551 (sec)

\begin{align*} y^{\prime \prime }+y^{\prime } \sin \left (x \right )+{y^{\prime }}^{2}&=0 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (x \right )+u \left (x \right ) \sin \left (x \right )+u \left (x \right )^{2} = 0 \end{align*}

Which is now solved for \(u(x)\) as first order ode.

Entering first order ode bernoulli solver In canonical form, the ODE is

\begin{align*} u' &= F(x,u)\\ &= -u \sin \left (x \right )-u^{2} \end{align*}

This is a Bernoulli ODE.

\[ u' = \left (-\sin \left (x \right )\right ) u \left (x \right ) + \left (-1\right )u^{2} \tag {1} \]

The standard Bernoulli ODE has the form

\[ u' = f_0(x)u+f_1(x)u^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-\sin \left (x \right )\\ f_1 &=-1 \end{align*}

The first step is to divide the above equation by \(u^n \) which gives

\[ \frac {u'}{u^n} = f_0(x) u^{1-n} +f_1(x) \tag {3} \]

The next step is use the substitution \(v = u^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(u(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=-\sin \left (x \right )\\ f_1(x)&=-1\\ n &=2 \end{align*}

Dividing both sides of ODE (1) by \(u^n=u^{2}\) gives

\begin{align*} u'\frac {1}{u^{2}} &= -\frac {\sin \left (x \right )}{u} -1 \tag {4} \end{align*}

Let

\begin{align*} v &= u^{1-n} \\ &= \frac {1}{u} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= -\frac {1}{u^{2}}u' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -v^{\prime }\left (x \right )&= -\sin \left (x \right ) v \left (x \right )-1\\ v' &= \sin \left (x \right ) v +1 \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (x \right ) + q(x)v \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\sin \left (x \right )\\ p(x) &=1 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\sin \left (x \right )d x}\\ &= {\mathrm e}^{\cos \left (x \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \mu \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{\cos \left (x \right )}\right ) &= {\mathrm e}^{\cos \left (x \right )}\\ \mathrm {d} \left (v \,{\mathrm e}^{\cos \left (x \right )}\right ) &= {\mathrm e}^{\cos \left (x \right )}\mathrm {d} x \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{\cos \left (x \right )}&= \int {{\mathrm e}^{\cos \left (x \right )} \,dx} \\ &=\int {\mathrm e}^{\cos \left (x \right )}d x + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\cos \left (x \right )}\) gives the final solution

\[ v \left (x \right ) = {\mathrm e}^{-\cos \left (x \right )} \left (\int {\mathrm e}^{\cos \left (x \right )}d x +c_1 \right ) \]

The substitution \(v = u^{1-n}\) is now used to convert the above solution back to \(u \left (x \right )\) which results in

\[ \frac {1}{u \left (x \right )} = {\mathrm e}^{-\cos \left (x \right )} \left (\int {\mathrm e}^{\cos \left (x \right )}d x +c_1 \right ) \]

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1} \\ \end{align*}

In summary, these are the solution found for \(y\)

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1} \\ \end{align*}

For solution \(u \left (x \right ) = \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1}\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {\frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1}\, dx}\\ y &= \ln \left (\int {\mathrm e}^{\cos \left (x \right )}d x +c_1 \right ) + c_2 \end{align*}

\begin{align*} y&= \int \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1}d x +c_2 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= \int \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1}d x +c_2 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \int \frac {{\mathrm e}^{\cos \left (x \right )}}{\int {\mathrm e}^{\cos \left (x \right )}d x +c_1}d x +c_2 \\ \end{align*}

2.2.10.2 second order nonlinear solved by mainardi lioville method

0.273 (sec)

\begin{align*} y^{\prime \prime }+y^{\prime } \sin \left (x \right )+{y^{\prime }}^{2}&=0 \\ \end{align*}

Entering second order nonlinear solved by mainardi lioville method solverThe 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) &= \sin \left (x \right )\\ g(y) &= 1 \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 \(g=1\) and \(f=\sin \left (x \right )\), then

\begin{align*} \int -g d y &= \int \left (-1\right )d y\\ &= -y\\ \int -f d x &= \int -\sin \left (x \right )d x\\ &= \cos \left (x \right ) \end{align*}

Substituting the above into Eq(6A) gives

\[ y^{\prime } = c_2 \,{\mathrm e}^{-y} {\mathrm e}^{\cos \left (x \right )} \]

Which is now solved as first order separable ode. The ode

\begin{equation} y^{\prime } = c_2 \,{\mathrm e}^{-y} {\mathrm e}^{\cos \left (x \right )} \end{equation}

is separable as it can be written as

\begin{align*} y^{\prime }&= c_2 \,{\mathrm e}^{-y} {\mathrm e}^{\cos \left (x \right )}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= {\mathrm e}^{\cos \left (x \right )} c_2\\ g(y) &= {\mathrm e}^{-y} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx} \\ \int { {\mathrm e}^{y}\,dy} &= \int { {\mathrm e}^{\cos \left (x \right )} c_2 \,dx} \\ \end{align*}

\[ {\mathrm e}^{y}=\int {\mathrm e}^{\cos \left (x \right )} c_2 d x +c_3 \]

Summary of solutions found

\begin{align*} {\mathrm e}^{y} &= \int {\mathrm e}^{\cos \left (x \right )} c_2 d x +c_3 \\ \end{align*}

2.2.10.3 SSolved using second order ode arccos transformation

0.925 (sec)

\begin{align*} y^{\prime \prime }+y^{\prime } \sin \left (x \right )+{y^{\prime }}^{2}&=0 \\ \end{align*}

Applying change of variable \(x = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ \left (-\tau ^{2}+1\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-\left (\left (\tau ^{2}-1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )-\tau ^{2}+\tau +1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right ) = 0 \]

Which is now solved for \(y \left (\tau \right )\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y \left (\tau \right )\). Let

\begin{align*} u(\tau ) &= \frac {d}{d \tau }y \left (\tau \right ) \end{align*}

Then

\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) \end{align*}

Hence the ode becomes

\begin{align*} \left (-\tau ^{2}+1\right ) \left (\frac {d}{d \tau }u \left (\tau \right )\right )-\left (\left (\tau ^{2}-1\right ) u \left (\tau \right )-\tau ^{2}+\tau +1\right ) u \left (\tau \right ) = 0 \end{align*}

Which is now solved for \(u(\tau )\) as first order ode.

Entering first order ode bernoulli solver In canonical form, the ODE is

\begin{align*} u' &= F(\tau ,u)\\ &= -\frac {u \left (\tau ^{2} u -\tau ^{2}+\tau -u +1\right )}{\tau ^{2}-1} \end{align*}

This is a Bernoulli ODE.

\[ u' = \left (-\frac {-\tau ^{2}+\tau +1}{\tau ^{2}-1}\right ) u \left (\tau \right ) + \left (-1\right )u^{2} \tag {1} \]

The standard Bernoulli ODE has the form

\[ u' = f_0(\tau )u+f_1(\tau )u^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-\frac {-\tau ^{2}+\tau +1}{\tau ^{2}-1}\\ f_1 &=-1 \end{align*}

The first step is to divide the above equation by \(u^n \) which gives

\[ \frac {u'}{u^n} = f_0(\tau ) u^{1-n} +f_1(\tau ) \tag {3} \]

The next step is use the substitution \(v = u^{1-n}\) in equation (3) which generates a new ODE in \(v \left (\tau \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(u(\tau )\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(\tau )&=-\frac {-\tau ^{2}+\tau +1}{\tau ^{2}-1}\\ f_1(\tau )&=-1\\ n &=2 \end{align*}

Dividing both sides of ODE (1) by \(u^n=u^{2}\) gives

\begin{align*} u'\frac {1}{u^{2}} &= -\frac {-\tau ^{2}+\tau +1}{\left (\tau ^{2}-1\right ) u} -1 \tag {4} \end{align*}

Let

\begin{align*} v &= u^{1-n} \\ &= \frac {1}{u} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(\tau \) gives

\begin{align*} v' &= -\frac {1}{u^{2}}u' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -\frac {d}{d \tau }v \left (\tau \right )&= -\frac {\left (-\tau ^{2}+\tau +1\right ) v \left (\tau \right )}{\tau ^{2}-1}-1\\ v' &= \frac {\left (-\tau ^{2}+\tau +1\right ) v}{\tau ^{2}-1}+1 \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (\tau \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} \frac {d}{d \tau }v \left (\tau \right ) + q(\tau )v \left (\tau \right ) &= p(\tau ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\tau ) &=-\frac {-\tau ^{2}+\tau +1}{\tau ^{2}-1}\\ p(\tau ) &=1 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int -\frac {-\tau ^{2}+\tau +1}{\tau ^{2}-1}d \tau }\\ &= \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }}\left ( \mu v\right ) &= \mu \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (\frac {v \,{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right ) &= \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\\ \mathrm {d} \left (\frac {v \,{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right ) &= \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\mathrm {d} \tau \end{align*}

Integrating gives

\begin{align*} \frac {v \,{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}&= \int {\frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}} \,d\tau } \\ &=\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau + c_1 \end{align*}

Dividing throughout by the integrating factor \(\frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\) gives the final solution

\[ v \left (\tau \right ) = \sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-\tau } \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right ) \]

The substitution \(v = u^{1-n}\) is now used to convert the above solution back to \(u \left (\tau \right )\) which results in

\[ \frac {1}{u \left (\tau \right )} = \sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-\tau } \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right ) \]

Solving for \(u \left (\tau \right )\) gives

\begin{align*} u \left (\tau \right ) &= \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )} \\ \end{align*}

In summary, these are the solution found for \(y \left (\tau \right )\)

\begin{align*} u \left (\tau \right ) &= \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )} \\ \end{align*}

For solution \(u \left (\tau \right ) = \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )}\), since \(u=\frac {d}{d \tau }y \left (\tau \right )\) then we now have a new first order ode to solve which is

\begin{align*} \frac {d}{d \tau }y \left (\tau \right ) = \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(\frac {d}{d \tau }y \left (\tau \right )=f(\tau )\), then we only need to integrate \(f(\tau )\).

\begin{align*} \int {dy} &= \int {\frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )}\, d\tau }\\ y \left (\tau \right ) &= \ln \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right ) + c_2 \end{align*}

\begin{align*} y \left (\tau \right )&= \int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )}d \tau +c_2 \end{align*}

In summary, these are the solution found for \((y \left (\tau \right ))\)

\begin{align*} y \left (\tau \right ) &= \int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\int \frac {{\mathrm e}^{\tau }}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_1 \right )}d \tau +c_2 \\ \end{align*}

Applying change of variable \(\tau = \cos \left (x \right )\) to the solutions above gives

\begin{align*} y &= \int -\frac {{\mathrm e}^{\cos \left (x \right )} \sin \left (x \right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}\, \left (\int -\frac {{\mathrm e}^{\cos \left (x \right )} \sin \left (x \right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}d x +c_1 \right )}d x +c_2 \\ \end{align*}

2.2.10.4 ✓ Maple. Time used: 0.005 (sec). Leaf size: 14

ode:=diff(diff(y(x),x),x)+sin(x)*diff(y(x),x)+diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \ln \left (c_1 \int {\mathrm e}^{\cos \left (x \right )}d x +c_2 \right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville successful
 

2.2.10.5 ✓ Mathematica. Time used: 2.222 (sec). Leaf size: 63

ode=D[y[x],{x,2}]+Sin[x]*D[y[x],x]+(D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \int _1^x\frac {\exp \left (\int _1^{K[3]}-\sin (K[1])dK[1]\right )}{c_1-\int _1^{K[3]}-\exp \left (\int _1^{K[2]}-\sin (K[1])dK[1]\right )dK[2]}dK[3]+c_2 \end{align*}

2.2.10.6 ✓ Sympy. Time used: 82.216 (sec). Leaf size: 37

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(sin(x)*Derivative(y(x), x) + Derivative(y(x), x)**2 + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \left [ y{\left (x \right )} = C_{1} + \int \frac {e^{\cos {\left (x \right )}}}{C_{2} + \int e^{\cos {\left (x \right )}}\, dx}\, dx, \ y{\left (x \right )} = C_{1} + \int \frac {e^{\cos {\left (x \right )}}}{C_{2} + \int e^{\cos {\left (x \right )}}\, dx}\, dx\right ] \]