obtain the series solution of the differential equation: y^(II) +xy^I −y=x^2 +1 y(0)=1 and y^I (0)=2


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Question Number 131805 by Engr_Jidda last updated on 08/Feb/21

$o b t a i n t h e s e r i e s s o l u t i o n o f t h e d i f f e r e n t i a l$ $e q u a t i o n : y^{I I} + {x y}^{I} - y = x^{2} + 1$ $y (0) = 1 a n d y^{I} (0) = 2$
	Answered by physicstutes last updated on 08/Feb/21

	$y^{″} + {x y}^{'} - y = x^{2} + 1 . . . . . . . . . . . (i)$ $x = 0 when y = 1 and y^{'} = 2$ $\Rightarrow y^{″} = 2$ $differentiating equation (i) :$ $y^{‴} + {x y}^{″} = 2 x . . . . . . . . . . . . (i i)$ $\Rightarrow y^{‴} = 0$ $differentiating equation (i i) :$ $y^{i v} + {x y}^{‴} + y^{″} = 2 . . . . . . . . . (i i i)$ $\Rightarrow y^{i v} = 0$ $differentiating (i i i) :$ $y^{v} + {x y}^{i v} + 2 y^{‴} = 0$ $\Rightarrow y^{v} = 0$ $a typical solution should be : y = 1 + 2 x + x^{2} + . . . .$ $test : y = 1 + 2 x + x^{2}$ $y^{'} = 2 + 2 x$ $y^{″} = 2$ $\Rightarrow (2) + x (2 + 2 x) - (1 + 2 x + x^{2}) = x^{2} + 1$ $2 + 2 x + 2 x^{2} - 1 - 2 x - x^{2} = x^{2} + 1$ $x^{2} + 1 = x^{2} + 1$
	Commented by Engr_Jidda last updated on 08/Feb/21

	$t h a n k y o u s i r$
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