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Question Number 206142 by universe last updated on 07/Apr/24
let (d^2 y/dx^2 )+p(x)(dy/dx)+q(x)y=0 , x∈R where p(x) and q(x) are continuous function if y_1 = sinx−2cosx and y_2 = 2sinx +cosx are L.I (linearly independent) solution then ∣4p(0)+2q(1)∣ = ?
$$\:\:\:\:\mathrm{let}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{p}\left({x}\right)\frac{{dy}}{{dx}}+{q}\left({x}\right){y}=\mathrm{0}\:,\:{x}\in\mathbb{R}\:\mathrm{where}\: \\ $$$$\:\:\:\:{p}\left({x}\right)\:\mathrm{and}\:{q}\left({x}\right)\:\mathrm{are}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{if} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} =\:\mathrm{sin}{x}−\mathrm{2cos}{x}\:{and}\:{y}_{\mathrm{2}} \:=\:\mathrm{2sin}{x}\:+\mathrm{cos}{x} \\ $$$$\:\:\:\:\mathrm{are}\:{L}.{I}\:\left(\mathrm{linearly}\:\mathrm{independent}\right)\:\mathrm{solution} \\ $$$$\:\:\:\:\:\mathrm{then}\:\:\mid\mathrm{4}{p}\left(\mathrm{0}\right)+\mathrm{2}{q}\left(\mathrm{1}\right)\mid\:=\:?\:\:\: \\ $$
Answered by MaruMaru last updated on 08/Apr/24
easy we can write more briefly like this solution y(t)=c′_1 sin(t)+c′_2 cos(t) ∴y^((2)) (t)+y(t)=0 y^((2)) (t)=−c′_1 sin(t)−c′_2 cos(t) y(t)=c′_1 sin(t)+c′_2 cos(t) y^((2)) (t)+y(t)=0 ∴ 2
$$\mathrm{easy} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:\mathrm{more}\:\mathrm{briefly}\:\mathrm{like}\:\mathrm{this} \\ $$$$\mathrm{solution}\:{y}\left({t}\right)={c}'_{\mathrm{1}} \mathrm{sin}\left({t}\right)+{c}'_{\mathrm{2}} \mathrm{cos}\left({t}\right) \\ $$$$\therefore{y}^{\left(\mathrm{2}\right)} \left({t}\right)+{y}\left({t}\right)=\mathrm{0} \\ $$$${y}^{\left(\mathrm{2}\right)} \left({t}\right)=−{c}'_{\mathrm{1}} \mathrm{sin}\left({t}\right)−{c}'_{\mathrm{2}} \mathrm{cos}\left({t}\right) \\ $$$${y}\left({t}\right)={c}'_{\mathrm{1}} \mathrm{sin}\left({t}\right)+{c}'_{\mathrm{2}} \mathrm{cos}\left({t}\right) \\ $$$${y}^{\left(\mathrm{2}\right)} \left({t}\right)+{y}\left({t}\right)=\mathrm{0} \\ $$$$\therefore\:\mathrm{2} \\ $$

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