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{\bf Question}

\bigskip Find the general solution of the differential equation $$
t \frac{dx}{dt} = x + \frac{1}{2}t \sec ^2 \frac{x}{2t}$$

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{\bf Answer}

\bigskip
$\displaystyle t \frac{dx}{dt} = x + \frac{1}{2}t \sec ^2
\frac{x}{2t}$

Rewrite as $\displaystyle \frac{dx}{dt} = \frac{x}{t} +
\frac{1}{2} \sec ^2 \frac{x}{2t}$

This is of the form $\frac{dx}{dt} = f\left( \frac{x}{t} \right)$
So let $y = \frac{x}{t}$

$\Rightarrow \frac{dx}{dt} = t \frac{dy}{dt} + y = y + \frac{1}{2}
\sec ^2 \frac{1}{2}y$

So we can rewrite as $$t\frac{dy}{dt} = \frac{1}{2} \sec^2
\frac{1}{2}y$$ Cross Multiply $$\frac{dt}{t} = \frac{2}{\sec^2
\frac{1}{2}y}dy = 2 \cos ^2 \frac{1}{2}y \, dy$$ Now $2 \cos ^2
\frac{y}{2} = 1 + \cos y$ so the differential equation becomes
$$\int \frac{dt}{t} = \int \left( 1+ \cos y\right) \, dy$$
Integrating $$ \ln |t| = y + \sin y + \mathrm{constant}$$ $$t =
Ae^{y + \sin y}$$ with $A$ as a constant.



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