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A homogeneous differential equation is a differential equation of the form $$y'=f(\frac{y}{t})$$

By letting $y'=0$, we can solve for $f(\frac{y}{t})=0$.

This gives us $y = a\cdot t$ for some constant $a$.

I can conclude that, the line segments in the direction field that has $0$ slope forms a line.

However, using this approach does not give me the right answer.

Can you share your method in approaching this question.

Note: There are more than one correct choices.

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The slope of the direction field at a point $(t,y)$ is $f(y/t)$. Points in lines $y/t=a$ have the same slope.Looking at the pictures, (c) is a clear candidate, and although it is hard to tell, maybe also (a).

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    Same slope meaning it can be non-zero? Why non-zero? Can you point out the mistake in my reasoning posted above? – 2017-02-20
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    $y'=0$ will give you the horizontal "isoclines", which in fact lie on a straight line through the origin. But you must look at the whole direction field. That is, not only to $y'=0$, but to $y'=\text{constant}$. – 2017-02-20
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    $y' = \text{constant}$ $\implies f(y/t) = \text{constant}$ But $\frac{y}{t} = f^{-1}(\text{constant})$ only applies if $f$ is bijective. Can you explain in more detail please. – 2017-02-20
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    If $f$ is not bijective, them $f(y/t)=C$ will consist of more than one straight line. For instance $(y/t)^2=1$ consists of two lines: $y=t$ and $y=-t$. – 2017-02-20