Internal problem ID [8839]
Internal file name [OUTPUT/7774_Monday_June_06_2022_12_14_22_AM_36448317/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 505.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "bernoulli", "separable", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {x y^{2} {y^{\prime }}^{2}-2 y^{3} y^{\prime }+2 y^{2} x=x^{3}} \] The ode \begin {align*} x y^{2} {y^{\prime }}^{2}-2 y^{3} y^{\prime }+2 y^{2} x = x^{3} \end {align*}
is factored to \begin {align*} \left (y y^{\prime }-x \right ) \left (-x y y^{\prime }+2 y^{2}-x^{2}\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y y^{\prime }-x = 0\tag {1} \\ -x y y^{\prime }+2 y^{2}-x^{2} = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x}{y} \end {align*}
Where \(f(x)=x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {x \,d x} \\ \frac {y^{2}}{2}&=\frac {x^{2}}{2}+c_{1} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {x^{2}+2 c_{1}} \\ y &= -\sqrt {x^{2}+2 c_{1}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {x^{2}+2 c_{1}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {x^{2}+2 c_{1}} \] 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*} -x^{2} u \left (x \right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+2 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 {u^{2}-1}{u x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {u^{2}-1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-1}{u}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}-1}{u}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {\ln \left (u^{2}-1\right )}{2}&=\ln \left (x \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}-1} &= {\mathrm e}^{\ln \left (x \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}-1} &= c_{4} x \end {align*}
Which simplifies to \[ \sqrt {u \left (x \right )^{2}-1} = c_{4} {\mathrm e}^{c_{3}} x \] The solution is \[ \sqrt {u \left (x \right )^{2}-1} = c_{4} {\mathrm e}^{c_{3}} x \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \sqrt {\frac {y^{2}}{x^{2}}-1} = c_{4} {\mathrm e}^{c_{3}} x\\ \sqrt {\frac {-x^{2}+y^{2}}{x^{2}}} = c_{4} {\mathrm e}^{c_{3}} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {-x^{2}+y^{2}}{x^{2}}} &= c_{4} {\mathrm e}^{c_{3}} x \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {-x^{2}+y^{2}}{x^{2}}} = c_{4} {\mathrm e}^{c_{3}} x \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {-x^{2}+y^{2}}{x^{2}}} &= c_{4} {\mathrm e}^{c_{3}} x \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {-x^{2}+y^{2}}{x^{2}}} = c_{4} {\mathrm e}^{c_{3}} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{2} {y^{\prime }}^{2}-2 y^{3} y^{\prime }+2 y^{2} x =x^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {x}{y}, y^{\prime }=\frac {2 y^{2}-x^{2}}{x y}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y y^{\prime }=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y y^{\prime }d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y^{2}-x^{2}}{x y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli 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: 0.016 (sec). Leaf size: 52
dsolve(x*y(x)^2*diff(y(x),x)^2-2*y(x)^3*diff(y(x),x)+2*x*y(x)^2-x^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {x^{2}+c_{1}} \\ y \left (x \right ) &= -\sqrt {x^{2}+c_{1}} \\ y \left (x \right ) &= \sqrt {c_{1} x^{2}+1}\, x \\ y \left (x \right ) &= -\sqrt {c_{1} x^{2}+1}\, x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.561 (sec). Leaf size: 85
DSolve[-x^3 + 2*x*y[x]^2 - 2*y[x]^3*y'[x] + x*y[x]^2*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {x^2+2 c_1} \\ y(x)\to \sqrt {x^2+2 c_1} \\ y(x)\to -\sqrt {x^2+c_1 x^4} \\ y(x)\to \sqrt {x^2+c_1 x^4} \\ y(x)\to -x \\ y(x)\to x \\ \end{align*}