Homogenious

$\displaystyle v(t) = y'(t) \rightarrow 5y'' + 2y' = 0 = 5v' + 2v \Rightarrow \frac{d}{dt} (v(t)\large{e^{\frac{2t}{5}}}) =$ $\frac{2}{5}e^{\frac{2t}{5}}v(t) + v'(t)e^{\frac{2t}{5}} = e^{\frac{2t}{5}}(\frac{2}{5}v(t) + v'(t)) = 0 \Rightarrow$ $\displaystyle \int d(e^{\frac{2t}{5}}v(t)) = \int 0 \space dt \Rightarrow e^{\frac{2t}{5}}v(t) = k \Rightarrow$ $v(t) = y'(t) = \dfrac{dy}{dt} = k\large{e^{\frac{-2t}{5}}} \Rightarrow$ $\displaystyle \int dy = \int k\large{e^{\frac{-2t}{5}}} dt + K_2 \Rightarrow y(t) = K_1 \large{e^{\frac{-2t}{5}}} + K_2$

Characteristic equation:

5r^2 + 2r = 0

Which gives solutions:

r = 0, -2/5

Therefore solution to the DE is

Ce^(-2x/5) + K

Where C and K are constants