Internal problem ID [8776]
Internal file name [OUTPUT/7710_Sunday_June_05_2022_11_45_26_PM_70648277/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 441.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "separable", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {{y^{\prime }}^{2} x^{2}-4 x \left (y+2\right ) y^{\prime }+4 y \left (y+2\right )=0} \] Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {2 y+4+2 \sqrt {2 y+4}}{x} \tag {1} \\ y^{\prime }&=\frac {2 y+4-2 \sqrt {2 y+4}}{x} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {2 y +4+2 \sqrt {2 y +4}}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=2 y +4+2 \sqrt {2 y +4}\). Integrating both sides gives \begin{align*} \frac {1}{2 y +4+2 \sqrt {2 y +4}} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{2 y +4+2 \sqrt {2 y +4}} \,dy} &= \int {\frac {1}{x} \,d x} \\ \ln \left (2+\sqrt {2 y +4}\right )&=\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} 2+\sqrt {2 y +4} &= {\mathrm e}^{\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} 2+\sqrt {2 y +4} &= c_{2} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2} {\mathrm e}^{2 c_{1}} c_{2}^{2}}{2}-2 c_{2} x \,{\mathrm e}^{c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2} {\mathrm e}^{2 c_{1}} c_{2}^{2}}{2}-2 c_{2} x \,{\mathrm e}^{c_{1}} \] Verified OK.
Solving equation (2)
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {2 y +4-2 \sqrt {2 y +4}}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=2 y +4-2 \sqrt {2 y +4}\). Integrating both sides gives \begin{align*} \frac {1}{2 y +4-2 \sqrt {2 y +4}} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{2 y +4-2 \sqrt {2 y +4}} \,dy} &= \int {\frac {1}{x} \,d x} \\ \ln \left (\sqrt {2 y +4}-2\right )&=\ln \left (x \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {2 y +4}-2 &= {\mathrm e}^{\ln \left (x \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} \sqrt {2 y +4}-2 &= c_{4} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{2 c_{3}} x^{2} c_{4}^{2}}{2}+2 c_{4} {\mathrm e}^{c_{3}} x \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{2 c_{3}} x^{2} c_{4}^{2}}{2}+2 c_{4} {\mathrm e}^{c_{3}} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2} x^{2}-4 x \left (y+2\right ) y^{\prime }+4 y \left (y+2\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 {2 y+4-2 \sqrt {2 y+4}}{x}, y^{\prime }=\frac {2 y+4+2 \sqrt {2 y+4}}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y+4-2 \sqrt {2 y+4}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2 y+4-2 \sqrt {2 y+4}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{2 y+4-2 \sqrt {2 y+4}}d x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y\right )}{2}-\frac {\ln \left (2+\sqrt {2 y+4}\right )}{2}+\frac {\ln \left (-2+\sqrt {2 y+4}\right )}{2}=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2} \left (-2 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}+\frac {16 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}+8\right )}{-4 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-8}, y=\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2} \left (2 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-\frac {16 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}+8\right )}{4 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-8}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y+4+2 \sqrt {2 y+4}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2 y+4+2 \sqrt {2 y+4}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{2 y+4+2 \sqrt {2 y+4}}d x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y\right )}{2}+\frac {\ln \left (2+\sqrt {2 y+4}\right )}{2}-\frac {\ln \left (-2+\sqrt {2 y+4}\right )}{2}=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-2 x \sqrt {2}\, {\mathrm e}^{c_{1}}+\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}, y=2 x \sqrt {2}\, {\mathrm e}^{c_{1}}+\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\left \{y=\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2} \left (-2 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}+\frac {16 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}+8\right )}{-4 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-8}, y=\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2} \left (2 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-\frac {16 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}+8\right )}{4 \sqrt {2}\, \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}-\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}-8}\right \}, \left \{y=-2 x \sqrt {2}\, {\mathrm e}^{c_{1}}+\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}, y=2 x \sqrt {2}\, {\mathrm e}^{c_{1}}+\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: *** Sublevel 2 *** 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`
✓ Solution by Maple
Time used: 1.266 (sec). Leaf size: 137
dsolve(x^2*diff(y(x),x)^2-4*x*(y(x)+2)*diff(y(x),x)+4*y(x)*(y(x)+2) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -2 \\ y \left (x \right ) &= \frac {-2 \sqrt {2}\, \sqrt {c_{1} x^{2}}+x^{2}}{c_{1}} \\ y \left (x \right ) &= \frac {2 \sqrt {2}\, \sqrt {c_{1} x^{2}}+x^{2}}{c_{1}} \\ y \left (x \right ) &= \frac {\left (-8 c_{1}^{2}+x^{2}\right ) \left (-2 c_{1} \sqrt {2}+x \right ) x}{\left (-4 c_{1} \sqrt {2}\, x +8 c_{1}^{2}+x^{2}\right ) c_{1}^{2}} \\ y \left (x \right ) &= \frac {\left (-8 c_{1}^{2}+x^{2}\right ) \left (2 c_{1} \sqrt {2}+x \right ) x}{\left (4 c_{1} \sqrt {2}\, x +8 c_{1}^{2}+x^{2}\right ) c_{1}^{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.198 (sec). Leaf size: 69
DSolve[4*y[x]*(2 + y[x]) - 4*x*(2 + y[x])*y'[x] + x^2*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to e^{-c_1} x \left (x-2 \sqrt {2} e^{\frac {c_1}{2}}\right ) \\ y(x)\to e^{c_1} x^2-2 \sqrt {2} e^{\frac {c_1}{2}} x \\ y(x)\to -2 \\ y(x)\to 0 \\ \end{align*}