Internal problem ID [13492]
Internal file name [OUTPUT/12664_Friday_February_16_2024_12_04_48_AM_86645881/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.4 (b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {3 y y^{\prime \prime }-2 {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 8, y^{\prime }\left (0\right ) = 6] \end {align*}
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} 3 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 p \left (y \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {2 p}{3 y} \end {align*}
Where \(f(y)=\frac {2}{3 y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {2}{3 y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {2}{3 y} \,d y}\\ \ln \left (p \right )&=\frac {2 \ln \left (y \right )}{3}+c_{1}\\ p&={\mathrm e}^{\frac {2 \ln \left (y \right )}{3}+c_{1}}\\ &=c_{1} y^{\frac {2}{3}} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=8\) and \(p=6\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 6 = 4 c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=\frac {3 y^{\frac {2}{3}}}{2} \end {align*}
The constant \(c_{1} = {\frac {3}{2}}\) gives valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \frac {3 y^{\frac {2}{3}}}{2} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {2}{3 y^{\frac {2}{3}}}d y &= \int {dx}\\ 2 y^{\frac {1}{3}}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=8\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 4 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 4 \end {align*}
Trying the constant \begin {align*} c_{2} = 4 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} 2 y^{\frac {1}{3}} = x +4 \end {align*}
The constant \(c_{2} = 4\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{8} x^{3}+\frac {3}{2} x^{2}+6 x +8 \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{8} x^{3}+\frac {3}{2} x^{2}+6 x +8 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 11
dsolve([3*y(x)*diff(y(x),x$2)=2*diff(y(x),x)^2,y(0) = 8, D(y)(0) = 6],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x +4\right )^{3}}{8} \]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 14
DSolve[{3*y[x]*y''[x]==2*y'[x]^2,{y[0]==8,y'[0]==6}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{8} (x+4)^3 \]