solve y′′−y′+e^(2x) y = 0


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Question Number 112840 by bemath last updated on 10/Sep/20

$solve y^{″} - y^{'} + e^{2 x} y = 0$
	Answered by john santu last updated on 10/Sep/20
	$solve y′′−y′ +e^(2x) y = 0. substitute u = e^x { (((dy/dx) = e^x = u)),(((d^2 y/dx^2 ) = (d/dx)((dy/dx))= (d/dx)(u (dy/du))= u^2 (d^2 y/du^2 ) +u (dy/du))) :} Hence y′′−y′ + e^(2x) y = (u^2 (d^2 y/du^2 ) + u (dy/dx))−(u (dy/du))+u^2 y = 0 ⇒ u^2 (d^2 y/du^2 ) +u^2 y = 0 so : u^2 ((d^2 y/du^2 ) + y) = 0 because u = e^x , u ≠ 0 , so we can divide by u^2 ⇒ (d^2 y/du^2 ) + y = 0 homogenous solution λ^2 +1 = 0 ; λ = ± i y = C_1 cos u + C_2 sin u y = C_1 cos (e^x ) + C_2 sin (e^x ) ((JS)/(a math farmer))$
	$s o l v e y^{″} - y^{'} + e^{2 x} y = 0 .$ $s u b s t i t u t e u = e^{x}$ ${\begin{cases} \frac{d y}{d x} = e^{x} = u \\ \frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (u \frac{d y}{d u}) = u^{2} \frac{d^{2} y}{{d u}^{2}} + u \frac{d y}{d u} \end{cases}$ $H e n c e y^{″} - y^{'} + e^{2 x} y =$ $(u^{2} \frac{d^{2} y}{{d u}^{2}} + u \frac{d y}{d x}) - (u \frac{d y}{d u}) + u^{2} y = 0$ $\Rightarrow u^{2} \frac{d^{2} y}{{d u}^{2}} + u^{2} y = 0$ $s o : u^{2} (\frac{d^{2} y}{{d u}^{2}} + y) = 0$ $b e c a u s e u = e^{x}, u \neq 0, s o w e c a n$ $d i v i d e b y u^{2} \Rightarrow \frac{d^{2} y}{{d u}^{2}} + y = 0$ $h o m o g e n o u s s o l u t i o n$ $λ^{2} + 1 = 0; λ = \pm i$ $y = C_{1} \cos u + C_{2} \sin u$ $y = C_{1} \cos (e^{x}) + C_{2} \sin (e^{x})$ $\frac{J S}{a m a t h f a r m e r}$
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