2.1.15.1 Solved using first_order_ode_homog_A

Entering first order ode homog A solver

\[\begin {aligned} x^{\prime }&=\cos \left (\frac {x}{t}\right ) \end {aligned}\]

In canonical form, the ODE is

\begin{align*} x' &= F(t,x)\\ &= \cos \left (\frac {x}{t}\right )\tag {1} \end{align*}

An ode of the form \(x' = \frac {M(t,x)}{N(t,x)}\) is called homogeneous if the functions \(M(t,x)\) and \(N(t,x)\) are both homogeneous functions and of the same order. Recall that a function \(f(t,x)\) is homogeneous of order \(n\) if

\[ f(t^n t, t^n x)= t^n f(t,x) \]

In this case, it can be seen that both \(M=\cos \left (\frac {x}{t}\right )\) and \(N=1\) are both homogeneous and of the same order \(n=0\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {x}{t}\), or \(x=ut\). Hence

\[ \frac { \mathop {\mathrm {d}x}}{\mathop {\mathrm {d}t}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}t}}t + u \]

Applying the transformation \(x=ut\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}t}}t + u &= \cos \left (u \right )\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}t}} &= \frac {\cos \left (u \left (t \right )\right )-u \left (t \right )}{t} \end{align*}

Or

\[ u^{\prime }\left (t \right )-\frac {\cos \left (u \left (t \right )\right )-u \left (t \right )}{t} = 0 \]

Or

\[ u^{\prime }\left (t \right ) t +u \left (t \right )-\cos \left (u \left (t \right )\right ) = 0 \]

Which is now solved as separable in \(u \left (t \right )\).

The ode

\begin{align*} u^{\prime }\left (t \right ) = -\frac {u \left (t \right )-\cos \left (u \left (t \right )\right )}{t} \end{align*}

is separable as it can be written as

\begin{align*} u^{\prime }\left (t \right )&= -\frac {u \left (t \right )-\cos \left (u \left (t \right )\right )}{t}\\ &= f(t) g(u) \end{align*}

Where

\begin{align*} f(t) &= \frac {1}{t}\\ g(u) &= \cos \left (u \right )-u \end{align*}

Integrating gives

\begin{gather*} \begin {aligned} \int { \frac {1}{g(u)} \,du} &= \int { f(t) \,dt}\\ \int { \frac {1}{\cos \left (u \right )-u}\,du} &= \int { \frac {1}{t} \,dt} \end {aligned} \end{gather*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\begin{align*} \cos \left (u \right )-u&=0 \end{align*}

for \(u \left (t \right )\) gives

\begin{align*} u \left (t \right )&=\operatorname {RootOf}\left (-\cos \left (\textit {\_Z} \right )+\textit {\_Z} \right ) \end{align*}

Now we each such singular solution is checked if it verifies the ode itself and any initial conditions given. If it does not then the singular solution is removed.

The solution \(\operatorname {RootOf}\left (-\cos \left (\textit {\_Z} \right )+\textit {\_Z} \right )\) will not be used

Converting \(\int _{}^{u \left (t \right )}\frac {1}{\cos \left (\tau \right )-\tau }d \tau = \ln \left (t \right )+c_1\) back to \(x\) gives

\begin{align*} \int _{}^{\frac {x}{t}}\frac {1}{\cos \left (\tau \right )-\tau }d \tau = \ln \left (t \right )+c_1 \end{align*}

Direction field with Isoclines	Direction field

Summary of solutions found

\[ \int _{}^{\frac {x}{t}}\frac {1}{\cos \left (\tau \right )-\tau }d \tau = \ln \left (t \right )+c_1 \]