Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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

��

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com