Internal problem ID [13491]
Internal file name [OUTPUT/12663_Friday_February_16_2024_12_04_47_AM_69437583/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 (a).
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]]
\[ \boxed {y y^{\prime \prime }-{y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 5, y^{\prime }\left (0\right ) = 15] \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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-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 {p}{y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {1}{y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {1}{y} \,d y}\\ \ln \left (p \right )&=\ln \left (y \right )+c_{1}\\ p&={\mathrm e}^{\ln \left (y \right )+c_{1}}\\ &=c_{1} y \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=5\) and \(p=15\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 15 = 5 c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 3 \end {align*}
Trying the constant \begin {align*} c_{1} = 3 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=3 y \end {align*}
The constant \(c_{1} = 3\) 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 } = 3 y \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{3 y}d y &= \int {dx}\\ \frac {\ln \left (y \right )}{3}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=5\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (5\right )}{3} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \frac {\ln \left (5\right )}{3} \end {align*}
Trying the constant \begin {align*} c_{2} = \frac {\ln \left (5\right )}{3} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y \right )}{3} = x +\frac {\ln \left (5\right )}{3} \end {align*}
The constant \(c_{2} = \frac {\ln \left (5\right )}{3}\) 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 &= 5 \,{\mathrm e}^{3 x} \\ \end{align*}
Verification of solutions
\[ y = 5 \,{\mathrm e}^{3 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y \left (\frac {d}{d x}y^{\prime }\right )-{y^{\prime }}^{2}=0, y \left (0\right )=5, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=15\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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {1}{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 {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=y \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{{\mathrm e}^{c_{1}} x +c_{2}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{{\mathrm e}^{c_{1}} x +c_{2}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5={\mathrm e}^{c_{2}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} {\mathrm e}^{{\mathrm e}^{c_{1}} x +c_{2}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=15 \\ {} & {} & 15={\mathrm e}^{c_{1}} {\mathrm e}^{c_{2}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\ln \left (3\right ), c_{2} =\ln \left (5\right )\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=5 \,{\mathrm e}^{3 x} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=5 \,{\mathrm e}^{3 x} \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: 0.078 (sec). Leaf size: 10
dsolve([y(x)*diff(y(x),x$2)=diff(y(x),x)^2,y(0) = 5, D(y)(0) = 15],y(x), singsol=all)
\[ y \left (x \right ) = 5 \,{\mathrm e}^{3 x} \]
✓ Solution by Mathematica
Time used: 0.129 (sec). Leaf size: 12
DSolve[{y[x]*y''[x]==y'[x]^2,{y[0]==5,y'[0]==15}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 5 e^{3 x} \]