8.50 problem 13.9 (i)

Internal problem ID [13521]
Internal file name [OUTPUT/12693_Friday_February_16_2024_12_11_02_AM_57548871/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: 13.9 (i).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"

Maple gives the following as the ode type

[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], _Lagerstrom, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

\[ \boxed {y^{\prime \prime }-2 y y^{\prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

8.50.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-2 y y^{\prime }\right )d x &= 0 \\ -y^{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {y}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \tan \left (c_{2} \sqrt {c_{1}}\right ) \sqrt {c_{1}}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1+\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right )^{2}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{1} \sec \left (c_{2} \sqrt {c_{1}}\right )^{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=0 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \tan \left (x \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tan \left (x \right ) \\ \end{align*}

Figure 529: Solution plot

Verification of solutions

\[ y = \tan \left (x \right ) \] Verified OK.

8.50.2 Solving as second order ode missing x ode

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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 y p \left (y \right ) = 0 \end {align*}

Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { 2 y\,\mathop {\mathrm {d}y}}\\ &= y^{2}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(y=0\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 1 \end {align*}

Trying the constant \begin {align*} c_{1} = 1 \end {align*}

Substituting this in the general solution gives \begin {align*} p \left (y \right )&=y^{2}+1 \end {align*}

The constant \(c_{1} = 1\) 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 } = y^{2}+1 \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+1}d y &= \int {dx}\\ \arctan \left (y \right )&= x +c_{2} \end {align*}

Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}

The solutions are \begin {align*} c_{2} = 0 \end {align*}

Trying the constant \begin {align*} c_{2} = 0 \end {align*}

Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \arctan \left (y \right ) = x \end {align*}

The constant \(c_{2} = 0\) 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 &= \tan \left (x \right ) \\ \end{align*}

Figure 530: Solution plot

Verification of solutions

\[ y = \tan \left (x \right ) \] Verified OK.

8.50.3 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y^{\prime \prime }-2 y y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-2 y y^{\prime }\right )d x &= 0 \\ -y^{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {y}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \tan \left (c_{2} \sqrt {c_{1}}\right ) \sqrt {c_{1}}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1+\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right )^{2}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{1} \sec \left (c_{2} \sqrt {c_{1}}\right )^{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=0 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \tan \left (x \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tan \left (x \right ) \\ \end{align*}

Figure 531: Solution plot

Verification of solutions

\[ y = \tan \left (x \right ) \] Verified OK.

8.50.4 Solving as exact nonlinear second order ode ode

An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}

Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}

Looking at the the ode given we see that \begin {align*} a_2 &= 1\\ a_1 &= -2 y\\ a_0 &= 0 \end {align*}

Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1\,d y'} + \int {-2 y\,d y} + \int {0\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} -y^{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {y}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \tan \left (c_{2} \sqrt {c_{1}}\right ) \sqrt {c_{1}}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1+\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right )^{2}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{1} \sec \left (c_{2} \sqrt {c_{1}}\right )^{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=0 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \tan \left (x \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tan \left (x \right ) \\ \end{align*}

Figure 532: Solution plot

Verification of solutions

\[ y = \tan \left (x \right ) \] Verified OK.

8.50.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y^{\prime }-2 y y^{\prime }=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-2 y u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=2 y \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}u \left (y \right )\right )d y =\int 2 y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\arctan \left (\frac {y}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\tan \left (c_{2} \sqrt {c_{1}}\right ) \sqrt {c_{1}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} \left (1+\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right )^{2}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{1} \left (1+\tan \left (c_{2} \sqrt {c_{1}}\right )^{2}\right ) \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\tan \left (x \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\tan \left (x \right ) \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: 0.156 (sec). Leaf size: 6

dsolve([diff(y(x),x$2)=2*y(x)*diff(y(x),x),y(0) = 0, D(y)(0) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = \tan \left (x \right ) \]

✓ Solution by Mathematica

Time used: 10.217 (sec). Leaf size: 7

DSolve[{y''[x]==2*y[x]*y'[x],{y[0]==0,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \tan (x) \]