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Question Number 94903 by mathmax by abdo last updated on 21/May/20

solve y^(′′) +2y^′ +y =xe^(−x)

$$\mathrm{solve}\:\:\mathrm{y}^{''} \:+\mathrm{2y}^{'} \:+\mathrm{y}\:=\mathrm{xe}^{−\mathrm{x}} \\ $$

Answered by niroj last updated on 21/May/20

y^(′′) +2y^′ +y = x e^(−x) (D^2 +2D+1)y= xe^(−x) A.E., (m^2 +2m+1)=0 (m+1)^2 =0 , m= −1, −1 CF= (C_1 +C_2 x)e^(−x) PI = ((xe^(−x) )/((D+1)^2 )) = e^(−x) (( x)/((D−1+1)^2 )) = e^(−x) (( x)/D^2 ) = e^(−x) ∫(∫xdx)dx = e^(−x) ∫ (x^2 /2)dx = (e^(−x) /2)×(x^3 /3) = (1/6)e^(−x) x^3 y= CF+PI y = (C_1 +C_2 x)+(1/6)e^(−x) x^3 //.

$$\:\mathrm{y}^{''} \:+\mathrm{2y}^{'} \:+\mathrm{y}\:=\:\mathrm{x}\:\mathrm{e}^{−\mathrm{x}} \\ $$$$\:\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{2D}+\mathrm{1}\right)\mathrm{y}=\:\mathrm{xe}^{−\mathrm{x}} \\ $$$$\mathrm{A}.\mathrm{E}.,\:\:\left(\mathrm{m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\left(\mathrm{m}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:,\:\mathrm{m}=\:−\mathrm{1},\:−\mathrm{1} \\ $$$$\:\:\:\mathrm{CF}=\:\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} \mathrm{x}\right)\mathrm{e}^{−\mathrm{x}} \\ $$$$\:\:\mathrm{PI}\:=\:\:\frac{\mathrm{xe}^{−\mathrm{x}} }{\left(\mathrm{D}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\:\mathrm{e}^{−\mathrm{x}} \:\frac{\:\:\mathrm{x}}{\left(\mathrm{D}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\:\mathrm{e}^{−\mathrm{x}} \:\frac{\:\mathrm{x}}{\mathrm{D}^{\mathrm{2}} } \\ $$$$\:\:\:=\:\mathrm{e}^{−\mathrm{x}} \int\left(\int\mathrm{xdx}\right)\mathrm{dx} \\ $$$$\:\:=\:\mathrm{e}^{−\mathrm{x}} \int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{dx}\:=\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}×\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{e}^{−\mathrm{x}} \mathrm{x}^{\mathrm{3}} \\ $$$$\:\mathrm{y}=\:\mathrm{CF}+\mathrm{PI} \\ $$$$\:\mathrm{y}\:\:=\:\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} \mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{e}^{−\mathrm{x}} \:\mathrm{x}^{\mathrm{3}} \://. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 22/May/20

thank you sir.

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

Commented by niroj last updated on 22/May/20

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Answered by ElOuafi last updated on 22/May/20

let (E): y′′+2y′+y=xe^(−x) and (E_0 ):y′′+2y′+y=0 be its equation without second member.⇒(E_r ):r^2 +2r+1=0⇒r=−1 so y_1 =e^(−x) ∧ y_2 =xe^(−x) are solutions of (E_0 ) the general solution of (E_0 ) is: y_g_0 =λ_1 e^(−x) +λ_2 xe^(−x) with λ_1 ,λ_2 ∈R , let search now a particular solution of (E) under the forme: y_p =λ_1 (x)y_1 +λ_2 (x)y_2 witch λ_1 (x),λ_2 (x) satisfying λ_1 ^′ (x)e^(−x) +λ_2 ^′ (x).xe^(−x) =0∧ λ_1 ^′ (x).(e^(−x) )^′ +λ_2 ^′ (x).(xe^(−x) )′=xe^(−x) after solving system we find: λ_1 ^′ (x)=−x^2 ∧ λ_2 ^′ (x)=x ⇒λ_1 =−(x^3 /3) ∧ λ_2 =(x^2 /2) ⇒y_p =−(x^3 /3)y_1 +(x^2 /2)y_2 ⇔y_p =−(x^3 /3)e^(−x) +(x^3 /2)e^(−x) =(x^3 /6)e^(−x) the general solution of (E) is : y_g =y_p +y_g_0 y(x)=(x^3 /6)e^(−x) +λ_1 e^(−x) +λ_2 xe^(−x) ;(λ_1 ;λ_2 )∈R^2

$${let}\:\left({E}\right):\:{y}''+\mathrm{2}{y}'+{y}={xe}^{−{x}} \\ $$$${and}\:\left({E}_{\mathrm{0}} \right):{y}''+\mathrm{2}{y}'+{y}=\mathrm{0}\:{be}\:{its}\:{equation}\:{without}\: \\ $$$${second}\:{member}.\Rightarrow\left({E}_{{r}} \right):{r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}=\mathrm{0}\Rightarrow{r}=−\mathrm{1} \\ $$$${so}\:{y}_{\mathrm{1}} ={e}^{−{x}} \wedge\:{y}_{\mathrm{2}} ={xe}^{−{x}} \:{are}\:{solutions}\:{of}\:\left({E}_{\mathrm{0}} \right) \\ $$$${the}\:{general}\:{solution}\:{of}\:\left({E}_{\mathrm{0}} \right)\:{is}:\:{y}_{{g}_{\mathrm{0}} } =\lambda_{\mathrm{1}} {e}^{−{x}} +\lambda_{\mathrm{2}} {xe}^{−{x}} \\ $$$${with}\:\lambda_{\mathrm{1}} ,\lambda_{\mathrm{2}} \in\mathbb{R}\:,\:{let}\:{search}\:{now}\:{a}\:{particular}\:{solution} \\ $$$${of}\:\left({E}\right)\:{under}\:{the}\:{forme}:\:{y}_{{p}} =\lambda_{\mathrm{1}} \left({x}\right){y}_{\mathrm{1}} +\lambda_{\mathrm{2}} \left({x}\right){y}_{\mathrm{2}} \: \\ $$$${witch}\:\lambda_{\mathrm{1}} \left({x}\right),\lambda_{\mathrm{2}} \left({x}\right)\:{satisfying}\: \\ $$$$\lambda_{\mathrm{1}} ^{'} \left({x}\right){e}^{−{x}} +\lambda_{\mathrm{2}} ^{'} \left({x}\right).{xe}^{−{x}} =\mathrm{0}\wedge\:\lambda_{\mathrm{1}} ^{'} \left({x}\right).\left({e}^{−{x}} \right)^{'} +\lambda_{\mathrm{2}} ^{'} \left({x}\right).\left({xe}^{−{x}} \right)'={xe}^{−{x}} \\ $$$${after}\:{solving}\:{system}\:{we}\:{find}:\:\lambda_{\mathrm{1}} ^{'} \left({x}\right)=−{x}^{\mathrm{2}} \wedge\:\lambda_{\mathrm{2}} ^{'} \left({x}\right)={x} \\ $$$$\Rightarrow\lambda_{\mathrm{1}} =−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\wedge\:\lambda_{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{p}} =−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}{y}_{\mathrm{1}} +\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{y}_{\mathrm{2}} \Leftrightarrow{y}_{{p}} =−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}{e}^{−{x}} +\frac{{x}^{\mathrm{3}} }{\mathrm{2}}{e}^{−{x}} =\frac{{x}^{\mathrm{3}} }{\mathrm{6}}{e}^{−{x}} \\ $$$${the}\:{general}\:{solution}\:{of}\:\left({E}\right)\:{is}\:: \\ $$$${y}_{{g}} ={y}_{{p}} +{y}_{{g}_{\mathrm{0}} } \\ $$$${y}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{6}}{e}^{−{x}} +\lambda_{\mathrm{1}} {e}^{−{x}} +\lambda_{\mathrm{2}} {xe}^{−{x}} \:;\left(\lambda_{\mathrm{1}} ;\lambda_{\mathrm{2}} \right)\in\mathbb{R}^{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 22/May/20

thank you sir

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

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