2.2.22 problem 23

Internal problem ID [8821]
Book : Second order enumerated odes
Section : section 2
Problem number : 23
Date solved : Thursday, December 12, 2024 at 09:52:15 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }+\tan \left (x \right ) y^{\prime }+\cos \left (x \right )^{2} y&=0 \end{align*}

Solved as second order ode using change of variable on x method 2

Time used: 0.397 (sec)

In normal form the ode

\begin{align*} y^{\prime \prime }+\tan \left (x \right ) y^{\prime }+\cos \left (x \right )^{2} y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\tan \left (x \right )\\ q \left (x \right )&=\cos \left (x \right )^{2} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \tan \left (x \right )d x}d x\\ &= \int e^{\ln \left (\cos \left (x \right )\right )} \,dx\\ &= \int \cos \left (x \right )d x\\ &= \sin \left (x \right )\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\cos \left (x \right )^{2}}{\cos \left (x \right )^{2}}\\ &= 1\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+y \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=1\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}+1 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=1\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (1\right )}\\ &= \pm i \end{align*}

Hence

\begin{align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*}

Since roots are complex conjugate of each others, then let the roots be

\[ \lambda _{1,2} = \alpha \pm i \beta \]

Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as

\[ y \left (\tau \right ) = e^{\alpha \tau } \left ( c_1 \cos (\beta \tau ) + c_2 \sin (\beta \tau ) \right ) \]

Which becomes

\[ y \left (\tau \right ) = e^{0}\left (c_1 \cos \left (\tau \right )+c_2 \sin \left (\tau \right )\right ) \]

Or

\[ y \left (\tau \right ) = c_1 \cos \left (\tau \right )+c_2 \sin \left (\tau \right ) \]

Will add steps showing solving for IC soon.

The above solution is now transformed back to \(y\) using (6) which results in

\[ y = c_1 \cos \left (\sin \left (x \right )\right )+c_2 \sin \left (\sin \left (x \right )\right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \cos \left (\sin \left (x \right )\right )+c_2 \sin \left (\sin \left (x \right )\right ) \\ \end{align*}

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   trying a quadrature 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
   Change of variables used: 
      [x = arcsin(t)] 
   Linear ODE actually solved: 
      (-2*t^2+2)*u(t)+(-2*t^2+2)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`
 

Maple dsolve solution

Solving time : 0.105 (sec)
Leaf size : 15

dsolve(diff(diff(y(x),x),x)+tan(x)*diff(y(x),x)+cos(x)^2*y(x) = 0, 
       y(x),singsol=all)
 

\[ y = c_{1} \sin \left (\sin \left (x \right )\right )+c_{2} \cos \left (\sin \left (x \right )\right ) \]

Mathematica DSolve solution

Solving time : 0.093 (sec)
Leaf size : 18

DSolve[{D[y[x],{x,2}]+Tan[x]*D[y[x],x]+Cos[x]^2*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_2 \sin (\sin (x))+c_1 \cos (\sin (x)) \]