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Prove that if $y' = f(t,y)$ is an Euler homogeneous equation and $y_{1}(t)$ is a solution, then $y(t) = \frac 1k y_{1}(kt)$ is also a solution for every non-zero $k \in \mathbb{R}$.

Now a differential equation $y' = f(t,y)$ is Euler homogeneous if $f(kt, ky) = f(t, y)$ for any $k \in \mathbb{R}$.

We also know that an Euler homogeneous equation can be transformed into a separable equation of the form $\frac{v'}{F(v) - v} = \frac{1}{t}$ via the substitution $v = y/t$, where $F(v) = f(t, y)$.

That is as much as I know. Thanks.

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    $\begingroup$ i'm getting that $y_1^{'}(x) = \frac {1} {k} f(x, y_1(x))$ should be true if we want $y(t) = \frac 1 k y_1(kt)$ to be a solution, which is true for $k=1$ only. $\endgroup$
    – tortue
    Commented Aug 13, 2018 at 10:41

1 Answer 1

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If $y_1(t)$ is a solution, then $y_1'(t) = f(t, y_1(t))$ and for $k \in \mathbb{R}-\{0\}$, we have $$\left(\dfrac1ky_1(kt)\right)'=y_1'(kt)=f(kt, y_1(kt)) = f(kt, k\dfrac1ky_1(kt))=f(t, \dfrac1ky_1(kt))$$

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