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sin-x-y-2cos-x-y-y-sin-x-e-x-

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Question Number 118442 by bramlexs22 last updated on 17/Oct/20
sin x.y′′ +2cos x. y′−y sin x = e^x
$$\:\:\mathrm{sin}\:{x}.{y}''\:+\mathrm{2cos}\:{x}.\:{y}'−{y}\:\mathrm{sin}\:{x}\:=\:{e}^{{x}} \\ $$$$ \\ $$
Answered by john santu last updated on 17/Oct/20
To solve it, we write it first as (sin x.y′′+cos x.y′ )+ (cos x.y′ −sin x)= e^x by the product rule , the first two term in parentheses are the derivative of sin x.y′ , while the second two are the derivative of cos x.y . Then the equation can be written as (d/(dx ))( sin x.y′+ cos x.y ) = e^x ⇒ sin x. y′ + cos x.y = e^x + C_1 ⇒(d/dx) (sin x.y ) = e^x + C_1 ⇒ sin x. y = e^x + C_1 x + C_2 ⇒ y = ((e^x + C_1 x + C_2 )/(sin x))
$$\:{To}\:{solve}\:{it},\:{we}\:{write}\:{it}\:{first}\:{as}\: \\ $$$$\left(\mathrm{sin}\:{x}.{y}''+\mathrm{cos}\:{x}.{y}'\:\right)+\:\left(\mathrm{cos}\:{x}.{y}'\:−\mathrm{sin}\:{x}\right)=\:{e}^{{x}} \\ $$$${by}\:{the}\:{product}\:{rule}\:,\:{the}\:{first}\:{two} \\ $$$${term}\:{in}\:{parentheses}\:{are}\:{the}\:{derivative} \\ $$$${of}\:\mathrm{sin}\:{x}.{y}'\:,\:{while}\:{the}\:{second}\:{two}\: \\ $$$${are}\:{the}\:{derivative}\:{of}\:\mathrm{cos}\:{x}.{y}\:.\:{Then} \\ $$$${the}\:{equation}\:{can}\:{be}\:{written}\:{as} \\ $$$$\frac{{d}}{{dx}\:}\left(\:\mathrm{sin}\:{x}.{y}'+\:\mathrm{cos}\:{x}.{y}\:\right)\:=\:{e}^{{x}} \: \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}.\:{y}'\:+\:\mathrm{cos}\:{x}.{y}\:=\:{e}^{{x}} \:+\:{C}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\:\left(\mathrm{sin}\:{x}.{y}\:\right)\:=\:{e}^{{x}} \:+\:{C}_{\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}.\:{y}\:=\:{e}^{{x}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} \\ $$$$\Rightarrow\:{y}\:=\:\frac{{e}^{{x}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} }{\mathrm{sin}\:{x}} \\ $$$$ \\ $$

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