Question-88181

Question Number 88181 by ubaydulla last updated on 08/Apr/20

Commented by mathmax by abdo last updated on 09/Apr/20

5) y^{^{'}} + y = e^{2 x} (h e) \to y^{^{'}} + y = 0 \Rightarrow \frac{y^{^{'}}}{y} = - 1 \Rightarrow l n ∣ y ∣= - x + c \Rightarrow

y = k e^{- x} m v c m e t h o d y^{^{'}} = k^{^{'}} e^{- x} - {k e}^{- x}

(e) \Rightarrow k^{^{'}} e^{- x} - {k e}^{- x} + {k e}^{- x} = e^{2 x} \Rightarrow k^{^{'}} = e^{3 x} \Rightarrow k (x) = \frac{1}{3} e^{3 x} + λ \Rightarrow

y (x) = (\frac{1}{3} e^{3 x} + λ) e^{- x} = \frac{1}{3} e^{2 x} + λ e^{- x}