Internal problem ID [6781]
Internal file name [OUTPUT/6028_Tuesday_July_26_2022_05_04_37_AM_9550685/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 94. Factoring the left member. EXERCISES
Page 309
Problem number: 15.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {\left (x^{2}+y^{2}\right )^{2} {y^{\prime }}^{2}-4 x^{2} y^{2}=0} \] The ode \begin {align*} \left (x^{2}+y^{2}\right )^{2} {y^{\prime }}^{2}-4 x^{2} y^{2} = 0 \end {align*}
is factored to \begin {align*} \left (y^{2} y^{\prime }+y^{\prime } x^{2}+2 y x \right ) \left (y^{2} y^{\prime }+y^{\prime } x^{2}-2 y x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{2} y^{\prime }+y^{\prime } x^{2}+2 y x = 0\tag {1} \\ y^{2} y^{\prime }+y^{\prime } x^{2}-2 y 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*} u \left (x \right )^{2} x^{2} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}+2 u \left (x \right ) x^{2} = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (u^{2}+3\right )}{x \left (u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u \left (u^{2}+3\right )}{u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u \left (u^{2}+3\right )}{u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u \left (u^{2}+3\right )}{u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (u^{3}+3 u \right )}{3}&=-\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \left (u^{3}+3 u \right )^{\frac {1}{3}} &= {\mathrm e}^{-\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} {\left (u \left (u^{2}+3\right )\right )}^{\frac {1}{3}} &= \frac {c_{3}}{x} \end {align*}
Which simplifies to \[ {\left (u \left (x \right ) \left (u \left (x \right )^{2}+3\right )\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] The solution is \[ {\left (u \left (x \right ) \left (u \left (x \right )^{2}+3\right )\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} {\left (\frac {y \left (\frac {y^{2}}{x^{2}}+3\right )}{x}\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x}\\ {\left (\frac {y \left (y^{2}+3 x^{2}\right )}{x^{3}}\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} {\left (\frac {y \left (y^{2}+3 x^{2}\right )}{x^{3}}\right )}^{\frac {1}{3}} &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \\ \end{align*}
Verification of solutions
\[ {\left (\frac {y \left (y^{2}+3 x^{2}\right )}{x^{3}}\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} {\left (\frac {y \left (y^{2}+3 x^{2}\right )}{x^{3}}\right )}^{\frac {1}{3}} &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \\ \end{align*}
Verification of solutions
\[ {\left (\frac {y \left (y^{2}+3 x^{2}\right )}{x^{3}}\right )}^{\frac {1}{3}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] 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*} u \left (x \right )^{2} x^{2} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}-2 u \left (x \right ) x^{2} = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u^{3}-u}{x \left (u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{3}-u}{u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{3}-u}{u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{3}-u}{u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ -\ln \left (u \right )+\ln \left (u^{2}-1\right )&=-\ln \left (x \right )+c_{5} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (u \right )+\ln \left (u^{2}-1\right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{5}} \end {align*}
Which simplifies to \begin {align*} \frac {u^{2}-1}{u} &= \frac {c_{6}}{x} \end {align*}
The solution is \[ \frac {u \left (x \right )^{2}-1}{u \left (x \right )} = \frac {c_{6}}{x} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {x \left (\frac {y^{2}}{x^{2}}-1\right )}{y} = \frac {c_{6}}{x}\\ \frac {y^{2}-x^{2}}{x y} = \frac {c_{6}}{x} \end {align*}
Which simplifies to \begin {align*} -\frac {\left (x -y\right ) \left (x +y\right )}{y} = c_{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (x -y\right ) \left (x +y\right )}{y} &= c_{6} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (x -y\right ) \left (x +y\right )}{y} = c_{6} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (x -y\right ) \left (x +y\right )}{y} &= c_{6} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (x -y\right ) \left (x +y\right )}{y} = c_{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+y^{2}\right )^{2} {y^{\prime }}^{2}-4 x^{2} y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {2 y x}{x^{2}+y^{2}}, y^{\prime }=\frac {2 y x}{x^{2}+y^{2}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {2 y x}{x^{2}+y^{2}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y x}{x^{2}+y^{2}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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: 0.141 (sec). Leaf size: 255
dsolve((x^2+y(x)^2)^2*diff(y(x),x)^2=4*x^2*y(x)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {1-\sqrt {4 x^{2} c_{1}^{2}+1}}{2 c_{1}} \\ y \left (x \right ) &= \frac {1+\sqrt {4 x^{2} c_{1}^{2}+1}}{2 c_{1}} \\ y \left (x \right ) &= -\frac {2 \left (c_{1} x^{2}-\frac {\left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {2}{3}}}{4}\right )}{\sqrt {c_{1}}\, \left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= -\frac {\left (1+i \sqrt {3}\right ) \left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {1}{3}}}{4 \sqrt {c_{1}}}-\frac {\left (i \sqrt {3}-1\right ) x^{2} \sqrt {c_{1}}}{\left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {4 i \sqrt {3}\, c_{1} x^{2}+i \sqrt {3}\, \left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {2}{3}}+4 c_{1} x^{2}-\left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {2}{3}}}{4 \left (4+4 \sqrt {4 c_{1}^{3} x^{6}+1}\right )^{\frac {1}{3}} \sqrt {c_{1}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 15.845 (sec). Leaf size: 345
DSolve[(x^2+y[x]^2)^2*(y'[x])^2==4*x^2*y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (-\sqrt {4 x^2+e^{2 c_1}}-e^{c_1}\right ) \\ y(x)\to \frac {1}{2} \left (\sqrt {4 x^2+e^{2 c_1}}-e^{c_1}\right ) \\ y(x)\to \frac {\sqrt [3]{\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}}}{\sqrt [3]{2}}-\frac {\sqrt [3]{2} x^2}{\sqrt [3]{\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}}} \\ y(x)\to \frac {i 2^{2/3} \left (\sqrt {3}+i\right ) \left (\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}\right ){}^{2/3}+\sqrt [3]{2} \left (2+2 i \sqrt {3}\right ) x^2}{4 \sqrt [3]{\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}}} \\ y(x)\to \frac {\left (1-i \sqrt {3}\right ) x^2}{2^{2/3} \sqrt [3]{\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}}}-\frac {\left (1+i \sqrt {3}\right ) \sqrt [3]{\sqrt {4 x^6+e^{6 c_1}}+e^{3 c_1}}}{2 \sqrt [3]{2}} \\ y(x)\to 0 \\ \end{align*}