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

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

$${obtain}\:{the}\:{series}\:{solution}\:{of}\:{the}\:{differential}\: \\ $$$${equation}:\:{y}^{{II}} +{xy}^{{I}} −{y}={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{1}\:{and}\:{y}^{{I}} \left(\mathrm{0}\right)=\mathrm{2} \\ $$

Answered by physicstutes last updated on 08/Feb/21

y′′ + xy′ −y = x^2 +1...........(i) x = 0 when y = 1 and y′ = 2 ⇒ y′′ = 2 differentiating equation (i): y′′′ + xy′′ = 2x............(ii) ⇒ y′′′ = 0 differentiating equation (ii): y^(iv) + xy′′′ + y′′ = 2 .........(iii) ⇒ y^(iv) = 0 differentiating (iii): y^v + xy^(iv) + 2y′′′ = 0 ⇒ y^v = 0 a typical solution should be: y = 1 + 2x + x^2 +.... test: y = 1 + 2x + x^2 y′ = 2 + 2x y′′ = 2 ⇒ (2) + x(2+2x)−(1+2x+x^2 ) = x^2 +1 2 + 2x + 2x^2 −1−2x−x^2 = x^2 +1 x^2 +1 = x^2 +1

$${y}''\:+\:{xy}'\:−{y}\:=\:{x}^{\mathrm{2}} +\mathrm{1}...........\left({i}\right) \\ $$$${x}\:=\:\mathrm{0}\:\mathrm{when}\:{y}\:=\:\mathrm{1}\:\mathrm{and}\:{y}'\:=\:\mathrm{2} \\ $$$$\Rightarrow\:{y}''\:=\:\mathrm{2} \\ $$$$\mathrm{differentiating}\:\mathrm{equation}\:\left({i}\right): \\ $$$$\:{y}'''\:+\:{xy}''\:=\:\mathrm{2}{x}............\left({ii}\right) \\ $$$$\Rightarrow\:{y}'''\:=\:\mathrm{0} \\ $$$$\mathrm{differentiating}\:\mathrm{equation}\:\left({ii}\right): \\ $$$${y}^{{iv}} +\:{xy}'''\:+\:{y}''\:=\:\mathrm{2}\:.........\left({iii}\right) \\ $$$$\Rightarrow\:{y}^{{iv}} \:=\:\mathrm{0} \\ $$$$\mathrm{differentiating}\:\left({iii}\right): \\ $$$$\:{y}^{{v}} \:+\:{xy}^{{iv}} +\:\mathrm{2}{y}'''\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{y}^{{v}} \:=\:\mathrm{0} \\ $$$$\mathrm{a}\:\mathrm{typical}\:\mathrm{solution}\:\mathrm{should}\:\mathrm{be}:\:\:{y}\:=\:\mathrm{1}\:+\:\mathrm{2}{x}\:+\:{x}^{\mathrm{2}} +.... \\ $$$$\mathrm{test}:\:{y}\:=\:\mathrm{1}\:+\:\mathrm{2}{x}\:+\:{x}^{\mathrm{2}} \\ $$$${y}'\:=\:\mathrm{2}\:+\:\mathrm{2}{x} \\ $$$${y}''\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)\:+\:{x}\left(\mathrm{2}+\mathrm{2}{x}\right)−\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)\:=\:{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}\:+\:\mathrm{2}{x}\:+\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} =\:{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{1}\:=\:{x}^{\mathrm{2}} +\mathrm{1}\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Commented by Engr_Jidda last updated on 08/Feb/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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