<s>[INST] Differential equation with initial value problem I am trying to solve the differential equation $$y'= \frac{1}{ty}$$ where $y(1)=2$. 
I have no idea how to solve this, and any help would be great!
 [/INST] \begin{align}
\frac{dy}{dt}&=\frac{1}{ty} \\
y \, dy &= \frac{dt}{t} & \text{separate the variables}\\
\int y \, dy &= \int \frac{dt}{t} \\
\frac{y^2}{2}&=\ln|t|+C &\text{integrate both sides} \\
y(t)&=\pm \sqrt{2\ln|t|+C} & \text{solve for $y=y(t)$}
\end{align}
Applying $y(1)=2$, we get
\begin{align}2&=\pm\sqrt{2\ln|1|+C} \\
4&=2\ln|1|+C & \text{square both sides} \\
4&=2\cdot0+C & \text{because } \ln|1|=0 \\
4&=C
\end{align}
Thus, $\boxed{y= \sqrt{2\ln|t|+4}}$
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