Internal problem ID [4159]
Internal file name [OUTPUT/3652_Sunday_June_05_2022_10_01_56_AM_45904658/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 31
Problem number: 924.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_linear]
\[ \boxed {4 x^{2} {y^{\prime }}^{2}-4 x y y^{\prime }+y^{2}=8 x^{3}} \] 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 {y+2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x} \tag {1} \\ y^{\prime }&=\frac {y-2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{2 x}\\ q(x) &=\sqrt {x}\, \sqrt {2} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{2 x} = \sqrt {x}\, \sqrt {2} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{2 x}d x} \\ &= \frac {1}{\sqrt {x}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\sqrt {x}\, \sqrt {2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{\sqrt {x}}\right ) &= \left (\frac {1}{\sqrt {x}}\right ) \left (\sqrt {x}\, \sqrt {2}\right )\\ \mathrm {d} \left (\frac {y}{\sqrt {x}}\right ) &= \sqrt {2}\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{\sqrt {x}} &= \int {\sqrt {2}\,\mathrm {d} x}\\ \frac {y}{\sqrt {x}} &= \sqrt {2}\, x + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{\sqrt {x}}\) results in \begin {align*} y &= \sqrt {2}\, x^{\frac {3}{2}}+c_{1} \sqrt {x} \end {align*}
which simplifies to \begin {align*} y &= \sqrt {x}\, \left (\sqrt {2}\, x +c_{1} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x}\, \left (\sqrt {2}\, x +c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = \sqrt {x}\, \left (\sqrt {2}\, x +c_{1} \right ) \] Verified OK.
Solving equation (2)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{2 x}\\ q(x) &=-\sqrt {x}\, \sqrt {2} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{2 x} = -\sqrt {x}\, \sqrt {2} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{2 x}d x} \\ &= \frac {1}{\sqrt {x}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (-\sqrt {x}\, \sqrt {2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{\sqrt {x}}\right ) &= \left (\frac {1}{\sqrt {x}}\right ) \left (-\sqrt {x}\, \sqrt {2}\right )\\ \mathrm {d} \left (\frac {y}{\sqrt {x}}\right ) &= \left (-\sqrt {2}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{\sqrt {x}} &= \int {-\sqrt {2}\,\mathrm {d} x}\\ \frac {y}{\sqrt {x}} &= -\sqrt {2}\, x + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{\sqrt {x}}\) results in \begin {align*} y &= -\sqrt {2}\, x^{\frac {3}{2}}+c_{2} \sqrt {x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\sqrt {2}\, x^{\frac {3}{2}}+c_{2} \sqrt {x} \\ \end{align*}
Verification of solutions
\[ y = -\sqrt {2}\, x^{\frac {3}{2}}+c_{2} \sqrt {x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} {y^{\prime }}^{2}-4 x y y^{\prime }+y^{2}=8 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 {y-2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x}, y^{\prime }=\frac {y+2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y-2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x} \\ {} & \circ & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{2 x}-\sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y}{2 x}=-\sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x}\right )=-\mu \left (x \right ) \sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=-\frac {\mu \left (x \right )}{2 x} \\ {} & \circ & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int -\mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1} \\ {} & \circ & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int -\mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int -\mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1}}{\mu \left (x \right )} \\ {} & \circ & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{\sqrt {x}} \\ {} & {} & y=\sqrt {x}\, \left (\int -\sqrt {2}d x +c_{1} \right ) \\ {} & \circ & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\sqrt {x}\, \left (-\sqrt {2}\, x +c_{1} \right ) \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y+2 \sqrt {2}\, x^{\frac {3}{2}}}{2 x} \\ {} & \circ & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{2 x}+\sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y}{2 x}=\sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x}\right )=\mu \left (x \right ) \sqrt {x}\, \sqrt {2} \\ {} & \circ & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=-\frac {\mu \left (x \right )}{2 x} \\ {} & \circ & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int \mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1} \\ {} & \circ & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right ) \sqrt {x}\, \sqrt {2}d x +c_{1}}{\mu \left (x \right )} \\ {} & \circ & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{\sqrt {x}} \\ {} & {} & y=\sqrt {x}\, \left (\int \sqrt {2}d x +c_{1} \right ) \\ {} & \circ & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\sqrt {x}\, \left (\sqrt {2}\, x +c_{1} \right ) \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\sqrt {x}\, \left (-\sqrt {2}\, x +c_{1} \right ), y=\sqrt {x}\, \left (\sqrt {2}\, x +c_{1} \right )\right \} \end {array} \]
Maple trace
`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 Successful isolation of dy/dx: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful ------------------- * Tackling next ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.079 (sec). Leaf size: 30
dsolve(4*x^2*diff(y(x),x)^2-4*x*y(x)*diff(y(x),x) = 8*x^3-y(x)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \left (-\sqrt {2}\, x +c_{1} \right ) \sqrt {x} \\ y \left (x \right ) &= \left (\sqrt {2}\, x +c_{1} \right ) \sqrt {x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.076 (sec). Leaf size: 42
DSolve[4 x^2 (y'[x])^2-4 x y[x] y'[x]==8 x^3 -y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \sqrt {x} \left (-\sqrt {2} x+c_1\right ) \\ y(x)\to \sqrt {x} \left (\sqrt {2} x+c_1\right ) \\ \end{align*}