Internal problem ID [13497]
Internal file name [OUTPUT/12669_Friday_February_16_2024_12_10_38_AM_48170922/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: 13.5 (a).
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {y^{\prime \prime }-4 x \sqrt {y^{\prime }}=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right )-4 x \sqrt {p \left (x \right )} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= 4 x \sqrt {p} \end {align*}
Where \(f(x)=4 x\) and \(g(p)=\sqrt {p}\). Integrating both sides gives \begin{align*} \frac {1}{\sqrt {p}} \,dp &= 4 x \,d x \\ \int { \frac {1}{\sqrt {p}} \,dp} &= \int {4 x \,d x} \\ 2 \sqrt {p}&=2 x^{2}+c_{1} \\ \end{align*} The solution is \[ 2 \sqrt {p \left (x \right )}-2 x^{2}-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} 2 \sqrt {y^{\prime }}-2 x^{2}-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x^{4}+c_{1} x^{2}+\frac {1}{4} c_{1}^{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {1}{5} x^{5}+\frac {1}{3} c_{1} x^{3}+\frac {1}{4} c_{1}^{2} x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{5} x^{5}+\frac {1}{3} c_{1} x^{3}+\frac {1}{4} c_{1}^{2} x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{5} x^{5}+\frac {1}{3} c_{1} x^{3}+\frac {1}{4} c_{1}^{2} x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-4 x \sqrt {y^{\prime }}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )-4 x \sqrt {u \left (x \right )}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=4 x \sqrt {u \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )}}=4 x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )}}d x =\int 4 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {u \left (x \right )}=2 x^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x^{4}+c_{1} x^{2}+\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x^{4}+c_{1} x^{2}+\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=x^{4}+c_{1} x^{2}+\frac {1}{4} c_{1}^{2} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \left (x^{4}+c_{1} x^{2}+\frac {1}{4} c_{1}^{2}\right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {1}{5} x^{5}+\frac {1}{3} c_{1} x^{3}+\frac {1}{4} c_{1}^{2} x +c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.219 (sec). Leaf size: 51
dsolve(diff(y(x),x$2)=4*x*sqrt(diff(y(x),x)),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= \frac {x^{5}}{5}-\frac {2 x^{3}}{3 c_{1}}+\frac {x}{c_{1}^{2}}+c_{2} \\ y \left (x \right ) &= \frac {x^{5}}{5}+\frac {2 x^{3}}{3 c_{1}}+\frac {x}{c_{1}^{2}}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.071 (sec). Leaf size: 33
DSolve[y''[x]==4*x*Sqrt[y'[x]],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^5}{5}+\frac {c_1 x^3}{3}+\frac {c_1{}^2 x}{4}+c_2 \]