Internal problem ID [4226]
Internal file name [OUTPUT/3719_Sunday_June_05_2022_10_25_30_AM_11571885/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 33
Problem number: 994.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, [_Abel, `2nd type`, `class A`]]
\[ \boxed {\left (x^{2}-4 y^{2}\right ) {y^{\prime }}^{2}+6 y y^{\prime } x +y^{2}=4 x^{2}} \] The ode \begin {align*} \left (x^{2}-4 y^{2}\right ) {y^{\prime }}^{2}+6 y y^{\prime } x +y^{2} = 4 x^{2} \end {align*}
is factored to \begin {align*} \left (2 y^{\prime } y+x y^{\prime }-y-2 x \right ) \left (2 y^{\prime } y-x y^{\prime }+y-2 x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} 2 y^{\prime } y+x y^{\prime }-y-2 x = 0\tag {1} \\ 2 y^{\prime } y-x y^{\prime }+y-2 x = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} 2 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x +x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-u \left (x \right ) x = 2 x \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 \left (u^{2}-1\right )}{x \left (2 u +1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{2}-1}{2 u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-1}{2 u +1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}-1}{2 u +1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ \frac {3 \ln \left (u -1\right )}{2}+\frac {\ln \left (u +1\right )}{2}&=-2 \ln \left (x \right )+c_{2} \\ \end{align*} The above can be written as \begin {align*} \frac {3 \ln \left (u -1\right )+\ln \left (u +1\right )}{2} &= -2 \ln \left (x \right )+c_{2}\\ 3 \ln \left (u -1\right )+\ln \left (u +1\right ) &= \left (2\right ) \left (-2 \ln \left (x \right )+c_{2}\right ) \\ &= -4 \ln \left (x \right )+2 c_{2} \end {align*}
Raising both side to exponential gives \begin {align*} {\mathrm e}^{3 \ln \left (u -1\right )+\ln \left (u +1\right )} &= {\mathrm e}^{-4 \ln \left (x \right )+2 c_{2}} \end {align*}
Which simplifies to \begin {align*} \left (u -1\right )^{3} \left (u +1\right ) &= \frac {2 c_{2}}{x^{4}}\\ &= \frac {c_{3}}{x^{4}} \end {align*}
Which simplifies to \[ u \left (x \right ) = \operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\frac {c_{3} {\mathrm e}^{2 c_{2}}}{x^{4}}+2 \textit {\_Z} -1\right ) \] Therefore the solution \(y\) is \begin {align*} y&=x u\\ &=x \operatorname {RootOf}\left (\textit {\_Z}^{4} x^{4}-2 \textit {\_Z}^{3} x^{4}+2 x^{4} \textit {\_Z} -x^{4}-c_{3} {\mathrm e}^{2 c_{2}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \operatorname {RootOf}\left (\textit {\_Z}^{4} x^{4}-2 \textit {\_Z}^{3} x^{4}+2 x^{4} \textit {\_Z} -x^{4}-c_{3} {\mathrm e}^{2 c_{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = x \operatorname {RootOf}\left (\textit {\_Z}^{4} x^{4}-2 \textit {\_Z}^{3} x^{4}+2 x^{4} \textit {\_Z} -x^{4}-c_{3} {\mathrm e}^{2 c_{2}}\right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \operatorname {RootOf}\left (\textit {\_Z}^{4} x^{4}-2 \textit {\_Z}^{3} x^{4}+2 x^{4} \textit {\_Z} -x^{4}-c_{3} {\mathrm e}^{2 c_{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = x \operatorname {RootOf}\left (\textit {\_Z}^{4} x^{4}-2 \textit {\_Z}^{3} x^{4}+2 x^{4} \textit {\_Z} -x^{4}-c_{3} {\mathrm e}^{2 c_{2}}\right ) \] Verified OK.
Solving ODE (2) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} 2 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x -x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+u \left (x \right ) x = 2 x \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 \left (u^{2}-1\right )}{x \left (2 u -1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{2}-1}{2 u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-1}{2 u -1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}-1}{2 u -1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ \ln \left (-u^{2}+1\right )+\operatorname {arctanh}\left (u \right )&=-2 \ln \left (x \right )+c_{5} \\ \end{align*} The solution is \[ \ln \left (-u \left (x \right )^{2}+1\right )+\operatorname {arctanh}\left (u \left (x \right )\right )+2 \ln \left (x \right )-c_{5} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} = 0\\ \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} &= 0 \\ \end{align*}
Verification of solutions
\[ \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} = 0 \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} &= 0 \\ \end{align*}
Verification of solutions
\[ \ln \left (1-\frac {y^{2}}{x^{2}}\right )+\operatorname {arctanh}\left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{5} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-4 y^{2}\right ) {y^{\prime }}^{2}+6 y y^{\prime } x +y^{2}=4 x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {2 x +y}{x +2 y}, y^{\prime }=-\frac {y-2 x}{-x +2 y}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 x +y}{x +2 y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y-2 x}{-x +2 y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 1.453 (sec). Leaf size: 93
dsolve((x^2-4*y(x)^2)*diff(y(x),x)^2+6*x*y(x)*diff(y(x),x)-4*x^2+y(x)^2 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x \left (-\operatorname {RootOf}\left (\textit {\_Z}^{16}+2 \textit {\_Z}^{4} c_{1} x^{4}-c_{1} x^{4}\right )^{4}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{16}+2 \textit {\_Z}^{4} c_{1} x^{4}-c_{1} x^{4}\right )^{4}} \\ y \left (x \right ) &= \frac {\frac {\operatorname {RootOf}\left (\textit {\_Z}^{16}-2 \textit {\_Z}^{4} c_{1} x^{4}-c_{1} x^{4}\right )^{12}}{c_{1}}-x^{4}}{x^{3}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.117 (sec). Leaf size: 3017
DSolve[(x^2-4 y[x]^2) (y'[x])^2 +6 x y[x] y'[x]-4 x^2+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
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