Internal problem ID [5865]
Internal file name [OUTPUT/5113_Sunday_June_05_2022_03_25_03_PM_89277202/index.tex]
Book: THEORY OF DIFFERENTIAL EQUATIONS IN ENGINEERING AND MECHANICS.
K.T. CHAU, CRC Press. Boca Raton, FL. 2018
Section: Chapter 3. Ordinary Differential Equations. Section 3.5 HIGHER ORDER ODE. Page
181
Problem number: Example 3.38.
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]]
\[ \boxed {x x^{\prime \prime }-{x^{\prime }}^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(x\) an independent variable. Using \begin {align*} x' &= p(x) \end {align*}
Then \begin {align*} x'' &= \frac {dp}{dt}\\ &= \frac {dx}{dt} \frac {dp}{dx}\\ &= p \frac {dp}{dx} \end {align*}
Hence the ode becomes \begin {align*} x p \left (x \right ) \left (\frac {d}{d x}p \left (x \right )\right )-p \left (x \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(x)\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {1}{x} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {\frac {1}{x} \,d x}\\ \ln \left (p \right )&=\ln \left (x \right )+c_{1}\\ p&={\mathrm e}^{\ln \left (x \right )+c_{1}}\\ &=c_{1} x \end {align*}
For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} x^{\prime } = c_{1} x \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{1} x}d x &= t +c_{2}\\ \frac {\ln \left (x \right )}{c_{1}}&=t +c_{2} \end {align*}
Solving for \(x\) gives these solutions \begin {align*} x_1&={\mathrm e}^{c_{1} c_{2} +c_{1} t}\\ &=c_{2} {\mathrm e}^{c_{1} t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= c_{2} {\mathrm e}^{c_{1} t} \\ \end{align*}
Verification of solutions
\[ x = c_{2} {\mathrm e}^{c_{1} t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x x^{\prime \prime }-{x^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (t \right )=x^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} x^{\prime \prime } \\ {} & {} & u^{\prime }\left (t \right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & x^{\prime } \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} x^{\prime }=u \left (x \right ),x^{\prime \prime }=u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & x u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )-u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {u \left (x \right )}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )}=\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )}d x =\int \frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (x \right )\right )=\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (x \right )=x^{\prime },x =x \\ {} & {} & x^{\prime }=x \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=x \,{\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{x}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{x}d t =\int {\mathrm e}^{c_{1}}d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (x\right )=t \,{\mathrm e}^{c_{1}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x={\mathrm e}^{t \,{\mathrm e}^{c_{1}}+c_{2}} \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: 14
dsolve(x(t)*diff(x(t),t$2)-diff(x(t),t)^2=0,x(t), singsol=all)
\begin{align*} x \left (t \right ) &= 0 \\ x \left (t \right ) &= {\mathrm e}^{c_{1} t} c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.108 (sec). Leaf size: 14
DSolve[x[t]*x''[t]-(x'[t])^2==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to c_2 e^{c_1 t} \]