Internal problem ID [13514]
Internal file name [OUTPUT/12686_Friday_February_16_2024_12_10_57_AM_78856885/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: 13.7 (c).
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 (1\right ) = 1, y^{\prime }\left (1\right ) = 9] \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 \(y=1\) and \(p=9\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 9 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 9 \end {align*}
Trying the constant \begin {align*} c_{1} = 9 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=9 y^{\frac {2}{3}} \end {align*}
The constant \(c_{1} = 9\) 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 } = 9 y^{\frac {2}{3}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{9 y^{\frac {2}{3}}}d y &= \int {dx}\\ \frac {y^{\frac {1}{3}}}{3}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=1\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{3}} = 1+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -{\frac {2}{3}} \end {align*}
Trying the constant \begin {align*} c_{2} = -{\frac {2}{3}} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \frac {y^{\frac {1}{3}}}{3} = x -\frac {2}{3} \end {align*}
The constant \(c_{2} = -{\frac {2}{3}}\) 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 &= 27 x^{3}-54 x^{2}+36 x -8 \\ \end{align*}
Verification of solutions
\[ y = 27 x^{3}-54 x^{2}+36 x -8 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 y \left (\frac {d}{d x}y^{\prime }\right )-2 {y^{\prime }}^{2}=0, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=9\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 3 y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-2 u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {2 u \left (y \right )}{3 y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {2}{3 y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {2}{3 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=\frac {2 \ln \left (y \right )}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{\frac {2 \ln \left (y \right )}{3}+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{\frac {2 \ln \left (y \right )}{3}+c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{\frac {2 \ln \left (y\right )}{3}+c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\frac {2 \ln \left (y\right )}{3}+c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{\frac {2 \ln \left (y\right )}{3}+c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{\frac {2 \ln \left (y\right )}{3}+c_{1}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {3 y}{{\mathrm e}^{\frac {2 \ln \left (y\right )}{3}+c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (c_{2}^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2} x +x^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}\right )^{\frac {3}{2}}}{27} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\left (c_{2}^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2} x +x^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}\right )^{\frac {3}{2}}}{27} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\frac {\left (c_{2}^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+\left ({\mathrm e}^{c_{1}}\right )^{2}\right )^{\frac {3}{2}}}{27} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {c_{2}^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2} x +x^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}\, \left (2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 x \left ({\mathrm e}^{c_{1}}\right )^{2}\right )}{18} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=9 \\ {} & {} & 9=\frac {\sqrt {c_{2}^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+\left ({\mathrm e}^{c_{1}}\right )^{2}}\, \left (2 c_{2} \left ({\mathrm e}^{c_{1}}\right )^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{2}\right )}{18} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 11
dsolve([3*y(x)*diff(y(x),x$2)=2*diff(y(x),x)^2,y(1) = 1, D(y)(1) = 9],y(x), singsol=all)
\[ y \left (x \right ) = \left (3 x -2\right )^{3} \]
✓ Solution by Mathematica
Time used: 0.033 (sec). Leaf size: 12
DSolve[{3*y[x]*y''[x]==2*y'[x]^2,{y[1]==1,y'[1]==9}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (3 x-2)^3 \]