Internal problem ID [13515]
Internal file name [OUTPUT/12687_Friday_February_16_2024_12_10_58_AM_7436523/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 (d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y y^{\prime \prime }+2 {y^{\prime }}^{2}-3 y y^{\prime }=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = {\frac {3}{4}}\right ] \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 )+\left (2 p \left (y \right )-3 y \right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\).
Entering Linear first order ODE solver. The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2}{y}d y} \\ &= y^{2} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu p\right ) &= \left (\mu \right ) \left (3\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (y^{2} p\right ) &= \left (y^{2}\right ) \left (3\right )\\ \mathrm {d} \left (y^{2} p\right ) &= \left (3 y^{2}\right )\, \mathrm {d} y \end {align*}
Integrating gives \begin {align*} y^{2} p &= \int {3 y^{2}\,\mathrm {d} y}\\ y^{2} p &= y^{3} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =y^{2}\) results in \begin {align*} p \left (y \right ) &= y +\frac {c_{1}}{y^{2}} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=2\) and \(p={\frac {3}{4}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {3}{4}} = \frac {c_{1}}{4}+2 \end {align*}
The solutions are \begin {align*} c_{1} = -5 \end {align*}
Trying the constant \begin {align*} c_{1} = -5 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=\frac {y^{3}-5}{y^{2}} \end {align*}
The constant \(c_{1} = -5\) 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 {y^{3}-5}{y^{2}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {y^{2}}{y^{3}-5}d y &= \int {dx}\\ \frac {\ln \left (y^{3}-5\right )}{3}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (3\right )}{3} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \frac {\ln \left (3\right )}{3} \end {align*}
Trying the constant \begin {align*} c_{2} = \frac {\ln \left (3\right )}{3} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y^{3}-5\right )}{3} = x +\frac {\ln \left (3\right )}{3} \end {align*}
The constant \(c_{2} = \frac {\ln \left (3\right )}{3}\) 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} \frac {\ln \left (y^{3}-5\right )}{3} &= x +\frac {\ln \left (3\right )}{3} \\ \end{align*}
Verification of solutions
\[ \frac {\ln \left (y^{3}-5\right )}{3} = x +\frac {\ln \left (3\right )}{3} \] Verified OK.
The ode has the Liouville form given by \begin {align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end {align*}
Where in this problem \begin {align*} f(x) &= -3\\ g(y) &= \frac {2}{y} \end {align*}
Dividing through by \(y^{\prime }\) then Eq (1A) becomes \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end {align*}
But the first term in Eq (2A) can be written as \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end {align*}
And the last term in Eq (2A) can be written as \begin {align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end {align*}
Substituting (3A,4A) back into (2A) gives \begin {align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end {align*}
Integrating the above w.r.t. \(x\) gives \begin {align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_{1} \end {align*}
Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives \begin {align*} y^{\prime } &= c_{2} e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end {align*}
Where \(c_{2}\) is a new arbitrary constant. But since \(g=\frac {2}{y}\) and \(f=-3\), then \begin {align*} \int -g d y &= \int -\frac {2}{y}d y\\ &= -2 \ln \left (y\right )\\ \int -f d x &= \int 3d x\\ &= 3 x \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = \frac {c_{2} {\mathrm e}^{3 x}}{y^{2}} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{2} {\mathrm e}^{3 x}}{y^{2}} \end {align*}
Where \(f(x)=c_{2} {\mathrm e}^{3 x}\) and \(g(y)=\frac {1}{y^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y^{2}}} \,dy &= c_{2} {\mathrm e}^{3 x} \,d x \\ \int { \frac {1}{\frac {1}{y^{2}}} \,dy} &= \int {c_{2} {\mathrm e}^{3 x} \,d x} \\ \frac {y^{3}}{3}&=\frac {c_{2} {\mathrm e}^{3 x}}{3}+c_{3} \\ \end{align*} The solution is \[ \frac {y^{3}}{3}-\frac {c_{2} {\mathrm e}^{3 x}}{3}-c_{3} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{3}}{3}-\frac {c_{2} {\mathrm e}^{3 x}}{3}-c_{3} = 0 \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 2\) and \(x = 0\) in the above gives \begin {align*} \frac {8}{3}-\frac {c_{2}}{3}-c_{3} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{2} {\mathrm e}^{3 x}}{\left (c_{2} {\mathrm e}^{3 x}+3 c_{3} \right )^{\frac {2}{3}}} \end {align*}
substituting \(y^{\prime } = {\frac {3}{4}}\) and \(x = 0\) in the above gives \begin {align*} {\frac {3}{4}} = \frac {c_{2}}{\left (c_{2} +3 c_{3} \right )^{\frac {2}{3}}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{2}&=3\\ c_{3}&={\frac {5}{3}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{3}}{3}-{\mathrm e}^{3 x}-\frac {5}{3} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{3}}{3}-{\mathrm e}^{3 x}-\frac {5}{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{3}}{3}-{\mathrm e}^{3 x}-\frac {5}{3} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime \prime }+\left (2 y^{\prime }-3 y\right ) y^{\prime }=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=\frac {3}{4}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \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 ),y^{\prime \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 )+\left (2 u \left (y \right )-3 y \right ) u \left (y \right )=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}{y} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=3-\frac {2 u \left (y \right )}{y} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )+\frac {2 u \left (y \right )}{y}=3 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (y \right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )+\frac {2 u \left (y \right )}{y}\right )=3 \mu \left (y \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )+\frac {2 u \left (y \right )}{y}\right )=\left (\frac {d}{d y}u \left (y \right )\right ) \mu \left (y \right )+u \left (y \right ) \left (\frac {d}{d y}\mu \left (y \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d y}\mu \left (y \right ) \\ {} & {} & \frac {d}{d y}\mu \left (y \right )=\frac {2 \mu \left (y \right )}{y} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (y \right )=y^{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right )\right )d y =\int 3 \mu \left (y \right )d y +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (y \right ) \mu \left (y \right )=\int 3 \mu \left (y \right )d y +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\int 3 \mu \left (y \right )d y +c_{1}}{\mu \left (y \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (y \right )=y^{2} \\ {} & {} & u \left (y \right )=\frac {\int 3 y^{2}d y +c_{1}}{y^{2}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & u \left (y \right )=\frac {y^{3}+c_{1}}{y^{2}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y^{3}+c_{1}}{y^{2}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {y^{3}+c_{1}}{y^{2}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}+c_{1}}{y^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y^{2}}{y^{3}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y^{2}}{y^{3}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y^{3}+c_{1} \right )}{3}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (-c_{1} +{\mathrm e}^{3 c_{2} +3 x}\right )^{\frac {1}{3}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\left (-c_{1} +{\mathrm e}^{3 c_{2} +3 x}\right )^{\frac {1}{3}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=\left (-c_{1} +{\mathrm e}^{3 c_{2}}\right )^{\frac {1}{3}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{3 c_{2} +3 x}}{\left (-c_{1} +{\mathrm e}^{3 c_{2} +3 x}\right )^{\frac {2}{3}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=\frac {3}{4} \\ {} & {} & \frac {3}{4}=\frac {{\mathrm e}^{3 c_{2}}}{\left (-c_{1} +{\mathrm e}^{3 c_{2}}\right )^{\frac {2}{3}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-5, c_{2} =\frac {\ln \left (3\right )}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (5+3 \,{\mathrm e}^{3 x}\right )^{\frac {1}{3}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (5+3 \,{\mathrm e}^{3 x}\right )^{\frac {1}{3}} \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: 0.125 (sec). Leaf size: 14
dsolve([y(x)*diff(y(x),x$2)+2*diff(y(x),x)^2=3*y(x)*diff(y(x),x),y(0) = 2, D(y)(0) = 3/4],y(x), singsol=all)
\[ y \left (x \right ) = \left (3 \,{\mathrm e}^{3 x}+5\right )^{\frac {1}{3}} \]
✓ Solution by Mathematica
Time used: 1.151 (sec). Leaf size: 18
DSolve[{y[x]*y''[x]+2*y'[x]^2==3*y[x]*y'[x],{y[0]==2,y'[0]==3/4}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt [3]{3 e^{3 x}+5} \]