1.492 problem 495

Internal problem ID [9474]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, linear first order
Problem number : 495
Date solved : Thursday, October 17, 2024 at 02:45:11 PM
CAS classification : [[_homogeneous, `class A`], _rational, _dAlembert]

Solve

\begin{align*} \left (y^{2}+\left (1-a \right ) x^{2}\right ) {y^{\prime }}^{2}+2 a x y y^{\prime }+\left (1-a \right ) y^{2}+x^{2}&=0 \end{align*}

1.492.1 Solved as first order homogeneous class A ode

Time used: 1.793 (sec)

Solving for \(y^{\prime }\) gives

\begin{align*} y^{\prime }&=-\frac {y a x +\sqrt {y^{4} a +2 y^{2} a \,x^{2}+a \,x^{4}-y^{4}-2 y^{2} x^{2}-x^{4}}}{-x^{2} a +y^{2}+x^{2}}\tag {1} \\ y^{\prime }&=-\frac {y a x -\sqrt {y^{4} a +2 y^{2} a \,x^{2}+a \,x^{4}-y^{4}-2 y^{2} x^{2}-x^{4}}}{-x^{2} a +y^{2}+x^{2}}\tag {2} \end{align*}

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {y a x +\sqrt {a \,x^{4}+2 y^{2} a \,x^{2}+y^{4} a -x^{4}-2 y^{2} x^{2}-y^{4}}}{-x^{2} a +x^{2}+y^{2}}\tag {1} \end{align*}

An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n y)= t^n f(x,y) \]

In this case, it can be seen that both \(M=y a x +\sqrt {a \,x^{4}+2 y^{2} a \,x^{2}+y^{4} a -x^{4}-2 y^{2} x^{2}-y^{4}}\) and \(N=x^{2} a -x^{2}-y^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence

\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(y=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \frac {u a +\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}}{-u^{2}+a -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right ) a +\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}}{-u \left (x \right )^{2}+a -1}-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right ) a +\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}}{-u \left (x \right )^{2}+a -1}-u \left (x \right )}{x} = 0 \]

Or

\[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x -u^{\prime }\left (x \right ) a x +u \left (x \right )^{3}+u^{\prime }\left (x \right ) x +u \left (x \right )+\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )} = 0 \]

Or

\[ \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+x \left (u \left (x \right )^{2}-a +1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{3}+u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode \(u^{\prime }\left (x \right ) = -\frac {u \left (x \right )^{3}+\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+u \left (x \right )}{x \left (u \left (x \right )^{2}-a +1\right )}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )^{3}+\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+u \left (x \right )}{x \left (u \left (x \right )^{2}-a +1\right )}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= \frac {u^{3}+\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}{u^{2}-a +1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {u^{2}-a +1}{u^{3}+\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}\,du} &= \int { -\frac {1}{x} \,dx}\\ \frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )-2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2}&=\ln \left (\frac {1}{x}\right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {u^{3}+\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}{u^{2}-a +1}=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-i\\ u \left (x \right )&=i \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )-2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2} = \ln \left (\frac {1}{x}\right )+c_1\\ u \left (x \right ) = -i\\ u \left (x \right ) = i \end{align*}

Converting \(\frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )-2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2} = \ln \left (\frac {1}{x}\right )+c_1\) back to \(y\) gives

\begin{align*} \frac {\ln \left (\frac {y^{2}+x^{2}}{x^{2}}\right ) \left (y^{2}+x^{2}\right )-2 \sqrt {\frac {\left (y^{2}+x^{2}\right )^{2} \left (a -1\right )}{x^{4}}}\, \arctan \left (\frac {y}{x}\right ) x^{2}}{2 y^{2}+2 x^{2}} = \ln \left (\frac {1}{x}\right )+c_1 \end{align*}

Converting \(u \left (x \right ) = -i\) back to \(y\) gives

\begin{align*} y = -i x \end{align*}

Converting \(u \left (x \right ) = i\) back to \(y\) gives

\begin{align*} y = i x \end{align*}

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {y a x -\sqrt {a \,x^{4}+2 y^{2} a \,x^{2}+y^{4} a -x^{4}-2 y^{2} x^{2}-y^{4}}}{-x^{2} a +x^{2}+y^{2}}\tag {1} \end{align*}

An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n y)= t^n f(x,y) \]

In this case, it can be seen that both \(M=y a x -\sqrt {a \,x^{4}+2 y^{2} a \,x^{2}+y^{4} a -x^{4}-2 y^{2} x^{2}-y^{4}}\) and \(N=x^{2} a -x^{2}-y^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence

\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(y=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \frac {u a -\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}}{-u^{2}+a -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right ) a -\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}}{-u \left (x \right )^{2}+a -1}-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right ) a -\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}}{-u \left (x \right )^{2}+a -1}-u \left (x \right )}{x} = 0 \]

Or

\[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x -u^{\prime }\left (x \right ) a x +u \left (x \right )^{3}+u^{\prime }\left (x \right ) x +u \left (x \right )-\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )} = 0 \]

Or

\[ -\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+x \left (u \left (x \right )^{2}-a +1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{3}+u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode \(u^{\prime }\left (x \right ) = -\frac {u \left (x \right )^{3}-\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+u \left (x \right )}{x \left (u \left (x \right )^{2}-a +1\right )}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )^{3}-\sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}+u \left (x \right )}{x \left (u \left (x \right )^{2}-a +1\right )}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= \frac {u^{3}-\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}{u^{2}-a +1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {u^{2}-a +1}{u^{3}-\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}\,du} &= \int { -\frac {1}{x} \,dx}\\ \frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )+2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2}&=\ln \left (\frac {1}{x}\right )+c_2 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {u^{3}-\sqrt {\left (u^{2}+1\right )^{2} \left (a -1\right )}+u}{u^{2}-a +1}=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-i\\ u \left (x \right )&=i \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )+2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2} = \ln \left (\frac {1}{x}\right )+c_2\\ u \left (x \right ) = -i\\ u \left (x \right ) = i \end{align*}

Converting \(\frac {\ln \left (u \left (x \right )^{2}+1\right ) \left (u \left (x \right )^{2}+1\right )+2 \sqrt {\left (u \left (x \right )^{2}+1\right )^{2} \left (a -1\right )}\, \arctan \left (u \left (x \right )\right )}{2 u \left (x \right )^{2}+2} = \ln \left (\frac {1}{x}\right )+c_2\) back to \(y\) gives

\begin{align*} \frac {\ln \left (\frac {y^{2}+x^{2}}{x^{2}}\right ) \left (y^{2}+x^{2}\right )+2 \sqrt {\frac {\left (y^{2}+x^{2}\right )^{2} \left (a -1\right )}{x^{4}}}\, \arctan \left (\frac {y}{x}\right ) x^{2}}{2 y^{2}+2 x^{2}} = \ln \left (\frac {1}{x}\right )+c_2 \end{align*}

Converting \(u \left (x \right ) = -i\) back to \(y\) gives

\begin{align*} y = -i x \end{align*}

Converting \(u \left (x \right ) = i\) back to \(y\) gives

\begin{align*} y = i x \end{align*}

1.492.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y \left (x \right )^{2}+\left (1-a \right ) x^{2}\right ) \left (\frac {d}{d x}y \left (x \right )\right )^{2}+2 a x y \left (x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (1-a \right ) y \left (x \right )^{2}+x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=-\frac {x a y \left (x \right )-\sqrt {a y \left (x \right )^{4}+2 x^{2} a y \left (x \right )^{2}+a \,x^{4}-y \left (x \right )^{4}-2 y \left (x \right )^{2} x^{2}-x^{4}}}{-a \,x^{2}+y \left (x \right )^{2}+x^{2}}, \frac {d}{d x}y \left (x \right )=-\frac {x a y \left (x \right )+\sqrt {a y \left (x \right )^{4}+2 x^{2} a y \left (x \right )^{2}+a \,x^{4}-y \left (x \right )^{4}-2 y \left (x \right )^{2} x^{2}-x^{4}}}{-a \,x^{2}+y \left (x \right )^{2}+x^{2}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {x a y \left (x \right )-\sqrt {a y \left (x \right )^{4}+2 x^{2} a y \left (x \right )^{2}+a \,x^{4}-y \left (x \right )^{4}-2 y \left (x \right )^{2} x^{2}-x^{4}}}{-a \,x^{2}+y \left (x \right )^{2}+x^{2}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {x a y \left (x \right )+\sqrt {a y \left (x \right )^{4}+2 x^{2} a y \left (x \right )^{2}+a \,x^{4}-y \left (x \right )^{4}-2 y \left (x \right )^{2} x^{2}-x^{4}}}{-a \,x^{2}+y \left (x \right )^{2}+x^{2}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

1.492.3 Maple trace

`Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying simple symmetries for implicit equations 
<- symmetries for implicit equations successful`
 

1.492.4 Maple dsolve solution

Solving time : 1.391 (sec)
Leaf size : 73

dsolve((y(x)^2+(-a+1)*x^2)*diff(y(x),x)^2+2*a*x*y(x)*diff(y(x),x)+(-a+1)*y(x)^2+x^2 = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= -i x \\ y &= i x \\ y &= \tan \left (\operatorname {RootOf}\left (-2 \textit {\_Z} \sqrt {a -1}-\ln \left (\sec \left (\textit {\_Z} \right )^{2} x^{2}\right )+2 c_{1} \right )\right ) x \\ y &= \tan \left (\operatorname {RootOf}\left (2 \textit {\_Z} \sqrt {a -1}-\ln \left (\sec \left (\textit {\_Z} \right )^{2} x^{2}\right )+2 c_{1} \right )\right ) x \\ \end{align*}

1.492.5 Mathematica DSolve solution

Solving time : 0.302 (sec)
Leaf size : 101

DSolve[{x^2 + (1 - a)*y[x]^2 + 2*a*x*y[x]*D[y[x],x] + ((1 - a)*x^2 + y[x]^2)*D[y[x],x]^2==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} \text {Solve}\left [\sqrt {a-1} \arctan \left (\frac {y(x)}{x}\right )-\frac {1}{2} \log \left (\frac {y(x)^2}{x^2}+1\right )&=\log (x)+c_1,y(x)\right ] \\ \text {Solve}\left [\sqrt {a-1} \arctan \left (\frac {y(x)}{x}\right )+\frac {1}{2} \log \left (\frac {y(x)^2}{x^2}+1\right )&=-\log (x)+c_1,y(x)\right ] \\ y(x)\to -i x \\ y(x)\to i x \\ \end{align*}