find the following differential equation by eliminating the arbritrary constant (1)y=Ae^x +Bcosx (2) xy=Ae^x +Be^(−x) +x^2


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Question Number 217245 by OmoloyeMichael last updated on 07/Mar/25

$f i n d t h e f o l l o w i n g d i f f e r e n t i a l e q u a t i o n$ $b y e l i m i n a t i n g t h e a r b r i t r a r y c o n s t a n t$ $(1) y = {A e}^{x} + B c o s x$ $(2) x y = {A e}^{x} + {B e}^{- x} + x^{2}$
	Answered by aleks041103 last updated on 07/Mar/25

	$f o r e x a m p l e :$ $(1) y = {A e}^{x} + B c o s x$ $y^{'} = {A e}^{x} - B s i n x$ $\Rightarrow y - y^{'} = B c o s x + B s i n x = z$ $z^{″} = - B c o s x - B s i n x = - z$ $\Rightarrow {(y - y^{'})}^{″} + (y - y^{'}) = 0$ $\Rightarrow y^{‴} - y^{″} + y^{'} - y = 0$ $(2) x y = {A e}^{x} + {B e}^{- x} + x^{2}$ $\Rightarrow z = x y - x^{2} = {A e}^{x} + {B e}^{- x}$ $z^{″} = {A e}^{x} + {B e}^{- x} = z$ $\Rightarrow {(x y - x^{2})}^{″} - (x y - x^{2}) = 0$ ${(x y)}^{″} - 2 - x y + x^{2} = 0$ ${(y + {x y}^{'})}^{'} - x y = 2 - x^{2}$ $y^{'} + y^{'} + {x y}^{″} - x y = 2 - x^{2}$ $\Rightarrow {x y}^{″} + 2 y^{'} - x y = 2 - x^{2}$
	Commented by aleks041103 last updated on 07/Mar/25

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