Internal problem ID [6783]
Internal file name [OUTPUT/6030_Tuesday_July_26_2022_05_04_42_AM_73131458/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: 17.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "bernoulli", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _Bernoulli]
\[ \boxed {x y \left (x^{2}+y^{2}\right ) \left ({y^{\prime }}^{2}-1\right )-y^{\prime } \left (x^{4}+x^{2} y^{2}+y^{4}\right )=0} \] The ode \begin {align*} x y \left (x^{2}+y^{2}\right ) \left ({y^{\prime }}^{2}-1\right )-y^{\prime } \left (x^{4}+x^{2} y^{2}+y^{4}\right ) = 0 \end {align*}
is factored to \begin {align*} \left (-y^{\prime } x y+y^{2}+x^{2}\right ) \left (y^{2} y^{\prime }+y^{\prime } x^{2}+y x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} -y^{\prime } x y+y^{2}+x^{2} = 0\tag {1} \\ y^{2} y^{\prime }+y^{\prime } x^{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*} -\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2} u \left (x \right )+u \left (x \right )^{2} x^{2} = -x^{2} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {1}{u x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {1}{u}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {u^{2}}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {u \left (x \right )^{2}}{2}-\ln \left (x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} = 0\\ \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} = 0 \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2 x^{2}}-\ln \left (x \right )-c_{2} = 0 \] 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}+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}+2\right )}{x \left (u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u \left (u^{2}+2\right )}{u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u \left (u^{2}+2\right )}{u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u \left (u^{2}+2\right )}{u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (u^{2}+2\right )}{4}+\frac {\ln \left (u \right )}{2}&=-\ln \left (x \right )+c_{4} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\frac {\ln \left (u^{2}+2\right )}{4}+\frac {\ln \left (u \right )}{2}} &= {\mathrm e}^{-\ln \left (x \right )+c_{4}} \end {align*}
Which simplifies to \begin {align*} \left (u^{2}+2\right )^{\frac {1}{4}} \sqrt {u} &= \frac {c_{5}}{x} \end {align*}
The solution is \[ \left (u \left (x \right )^{2}+2\right )^{\frac {1}{4}} \sqrt {u \left (x \right )} = \frac {c_{5}}{x} \] 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^{2}}{x^{2}}+2\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} = \frac {c_{5}}{x}\\ \left (\frac {y^{2}+2 x^{2}}{x^{2}}\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} = \frac {c_{5}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \left (\frac {y^{2}+2 x^{2}}{x^{2}}\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} &= \frac {c_{5}}{x} \\ \end{align*}
Verification of solutions
\[ \left (\frac {y^{2}+2 x^{2}}{x^{2}}\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} = \frac {c_{5}}{x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \left (\frac {y^{2}+2 x^{2}}{x^{2}}\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} &= \frac {c_{5}}{x} \\ \end{align*}
Verification of solutions
\[ \left (\frac {y^{2}+2 x^{2}}{x^{2}}\right )^{\frac {1}{4}} \sqrt {\frac {y}{x}} = \frac {c_{5}}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y \left (x^{2}+y^{2}\right ) \left ({y^{\prime }}^{2}-1\right )-y^{\prime } \left (x^{4}+x^{2} y^{2}+y^{4}\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {x^{2}+y^{2}}{y x}, y^{\prime }=-\frac {y x}{x^{2}+y^{2}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x^{2}+y^{2}}{y x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y x}{x^{2}+y^{2}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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 <- Bernoulli successful`
✓ Solution by Maple
Time used: 1.015 (sec). Leaf size: 248
dsolve(x*y(x)*(x^2+y(x)^2)*(diff(y(x),x)^2-1)=diff(y(x),x)*(x^4+x^2*y(x)^2+y(x)^4),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {x^{2} c_{1} \left (c_{1} x^{2}-\sqrt {c_{1}^{2} x^{4}+1}\right )}}{x \left (c_{1} x^{2}-\sqrt {c_{1}^{2} x^{4}+1}\right ) c_{1}} \\ y \left (x \right ) &= \frac {\sqrt {x^{2} c_{1} \left (c_{1} x^{2}+\sqrt {c_{1}^{2} x^{4}+1}\right )}}{x \left (c_{1} x^{2}+\sqrt {c_{1}^{2} x^{4}+1}\right ) c_{1}} \\ y \left (x \right ) &= \frac {\sqrt {x^{2} c_{1} \left (c_{1} x^{2}-\sqrt {c_{1}^{2} x^{4}+1}\right )}}{x \left (-c_{1} x^{2}+\sqrt {c_{1}^{2} x^{4}+1}\right ) c_{1}} \\ y \left (x \right ) &= -\frac {\sqrt {x^{2} c_{1} \left (c_{1} x^{2}+\sqrt {c_{1}^{2} x^{4}+1}\right )}}{x \left (c_{1} x^{2}+\sqrt {c_{1}^{2} x^{4}+1}\right ) c_{1}} \\ y \left (x \right ) &= \sqrt {2 \ln \left (x \right )+c_{1}}\, x \\ y \left (x \right ) &= -\sqrt {2 \ln \left (x \right )+c_{1}}\, x \\ \end{align*}
✓ Solution by Mathematica
Time used: 9.298 (sec). Leaf size: 248
DSolve[x*y[x]*(x^2+y[x]^2)*((y'[x])^2-1)==y'[x]*(x^4+x^2*y[x]^2+y[x]^4),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {-x^2-\sqrt {x^4+e^{4 c_1}}} \\ y(x)\to \sqrt {-x^2-\sqrt {x^4+e^{4 c_1}}} \\ y(x)\to -\sqrt {-x^2+\sqrt {x^4+e^{4 c_1}}} \\ y(x)\to \sqrt {-x^2+\sqrt {x^4+e^{4 c_1}}} \\ y(x)\to -x \sqrt {2 \log (x)+c_1} \\ y(x)\to x \sqrt {2 \log (x)+c_1} \\ y(x)\to -\sqrt {-\sqrt {x^4}-x^2} \\ y(x)\to \sqrt {-\sqrt {x^4}-x^2} \\ y(x)\to -\sqrt {\sqrt {x^4}-x^2} \\ y(x)\to \sqrt {\sqrt {x^4}-x^2} \\ \end{align*}