2.1.20.2 Solved using first_order_ode_separable

Entering first order ode separable solver

\[\begin {aligned} x^{\prime }+x \tan \left (t \right )&=0 \end {aligned}\]

The ode

\begin{align*} x^{\prime } = -x \tan \left (t \right ) \end{align*}

is separable as it can be written as

\begin{align*} x^{\prime }&= -x \tan \left (t \right )\\ &= f(t) g(x) \end{align*}

Where

\begin{align*} f(t) &= -\tan \left (t \right )\\ g(x) &= x \end{align*}

Integrating gives

\begin{gather*} \begin {aligned} \int { \frac {1}{g(x)} \,dx} &= \int { f(t) \,dt}\\ \int { \frac {1}{x}\,dx} &= \int { -\tan \left (t \right ) \,dt} \end {aligned} \end{gather*}

Which becomes

\begin{align*} \ln \left (x\right )&=\ln \left (\cos \left (t \right )\right )+c_1 \end{align*}

Taking the exponential of both sides the solution becomes

\begin{align*} x = c_1 \cos \left (t \right ) \end{align*}

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

\begin{align*} x&=0 \end{align*}

for \(x\) gives

\begin{align*} x&=0 \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.

Therefore the solutions found are

\begin{gather*} \begin {aligned} x &= c_1 \cos \left (t \right )\\ x &= 0\\ \end {aligned} \end{gather*}

Direction field with Isoclines	Direction field and Solutions plot

Summary of solutions found

\begin{align*} x&=c_1 \cos \left (t \right )\\ x&=0 \end{align*}