Internal problem ID [6854]
Internal file name [OUTPUT/6101_Friday_July_29_2022_03_09_17_AM_29950073/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 37.
ODE order: 2.
ODE degree: 1.
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]]
\[ \boxed {x^{4} y^{\prime \prime }-y^{\prime } \left (y^{\prime }+x^{3}\right )=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 2, y^{\prime }\left (1\right ) = 1] \end {align*}
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*} x^{4} p^{\prime }\left (x \right )+\left (-x^{3}-p \left (x \right )\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Using the change of variables \(p \left (x \right ) = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} x^{4} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+\left (-x^{3}-u \left (x \right ) x \right ) u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u^{2}}{x^{3}} \end {align*}
Where \(f(x)=\frac {1}{x^{3}}\) and \(g(u)=u^{2}\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}} \,du &= \frac {1}{x^{3}} \,d x \\ \int { \frac {1}{u^{2}} \,du} &= \int {\frac {1}{x^{3}} \,d x} \\ -\frac {1}{u}&=-\frac {1}{2 x^{2}}+c_{2} \\ \end{align*} The solution is \[ -\frac {1}{u \left (x \right )}+\frac {1}{2 x^{2}}-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {p \left (x \right )}{x}\) results in the solution for \(p \left (x \right )\) in implicit form \begin {align*} -\frac {x}{p \left (x \right )}+\frac {1}{2 x^{2}}-c_{2} = 0\\ -\frac {x}{p \left (x \right )}+\frac {1}{2 x^{2}}-c_{2} = 0 \end {align*}
Substituting initial conditions and solving for \(c_{2} \) gives \(c_{2} = -{\frac {1}{2}}\). Hence the solution becomes For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {x}{y^{\prime }}+\frac {1}{2 x^{2}}+\frac {1}{2} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {2 x^{3}}{x^{2}+1}\,\mathop {\mathrm {d}x}}\\ &= x^{2}-\ln \left (x^{2}+1\right )+c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = 1-\ln \left (2\right )+c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 1+\ln \left (2\right ) \end {align*}
Trying the constant \begin {align*} c_{3} = 1+\ln \left (2\right ) \end {align*}
Substituting this in the general solution gives \begin {align*} y&=x^{2}-\ln \left (x^{2}+1\right )+1+\ln \left (2\right ) \end {align*}
The constant \(c_{3} = 1+\ln \left (2\right )\) 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 &= x^{2}-\ln \left (x^{2}+1\right )+1+\ln \left (2\right ) \\ \end{align*}
Verification of solutions
\[ y = x^{2}-\ln \left (x^{2}+1\right )+1+\ln \left (2\right ) \] Verified OK.
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 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = _b(_a)*(_a^3+_b(_a))/_a^4, _b(_a), HINT = [[_a, 3*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 3*_b]
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 25
dsolve([x^4*diff(y(x),x$2)=diff(y(x),x)*(diff(y(x),x)+x^3),y(1) = 2, D(y)(1) = 1],y(x), singsol=all)
\[ y \left (x \right ) = x^{2}-\ln \left (-x^{2}-1\right )+1+\ln \left (2\right )+i \pi \]
✓ Solution by Mathematica
Time used: 0.929 (sec). Leaf size: 20
DSolve[{x^4*y''[x]==y'[x]*(y'[x]+x^3),{y[1]==2,y'[1]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^2-\log \left (x^2+1\right )+1+\log (2) \]