y-y-7-0-

Question Number 111644 by weltr last updated on 04/Sep/20

$${y}'+{y}+\mathrm{7}=\mathrm{0} \\ $$

Commented by mohammad17 last updated on 04/Sep/20

(D+1)y=−7 r+1=0⇒r=−1 y_h =C e^(−x) let:y_p =a⇒y_p ^′ =0 (0+a)=−7⇒a=−7 y=y_h +y_p y= C e^(−x) −7 by:≪mohammad≫

$$ \\ $$$$\left({D}+\mathrm{1}\right){y}=−\mathrm{7} \\ $$$$ \\ $$$${r}+\mathrm{1}=\mathrm{0}\Rightarrow{r}=−\mathrm{1} \\ $$$$ \\ $$$${y}_{{h}} ={C}\:{e}^{−{x}} \\ $$$$ \\ $$$${let}:{y}_{{p}} ={a}\Rightarrow{y}_{{p}} ^{'} =\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{0}+{a}\right)=−\mathrm{7}\Rightarrow{a}=−\mathrm{7} \\ $$$$ \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$$$ \\ $$$${y}=\:{C}\:{e}^{−{x}} −\mathrm{7} \\ $$$$ \\ $$$${by}:\ll{mohammad}\gg \\ $$

Commented by weltr last updated on 04/Sep/20

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

Commented by mohammad17 last updated on 04/Sep/20

method 2 (D+1)y=−7 (r+1)=0⇒r=−1 y_h =C e^(−x) let: y_p = ((1/(1+D)))(−7) y_p = {1−D+....}(−7) y_p =−7+0+...⇒y_p =−7 y=y_h +y_p y=C e^(−x) −7 ≪mohammad≫

$${method}\:\mathrm{2} \\ $$$$ \\ $$$$\left({D}+\mathrm{1}\right){y}=−\mathrm{7} \\ $$$$ \\ $$$$\left({r}+\mathrm{1}\right)=\mathrm{0}\Rightarrow{r}=−\mathrm{1} \\ $$$$ \\ $$$${y}_{{h}} ={C}\:{e}^{−{x}} \\ $$$$ \\ $$$${let}:\:{y}_{{p}} =\:\left(\frac{\mathrm{1}}{\mathrm{1}+{D}}\right)\left(−\mathrm{7}\right) \\ $$$$ \\ $$$${y}_{{p}} =\:\left\{\mathrm{1}−{D}+….\right\}\left(−\mathrm{7}\right) \\ $$$$ \\ $$$${y}_{{p}} =−\mathrm{7}+\mathrm{0}+…\Rightarrow{y}_{{p}} =−\mathrm{7} \\ $$$$ \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$$$ \\ $$$${y}={C}\:{e}^{−{x}} −\mathrm{7} \\ $$$$ \\ $$$$\ll{mohammad}\gg \\ $$

Commented by mohammad17 last updated on 04/Sep/20

$${you}\:{are}\:{welcome} \\ $$

Answered by mathmax by abdo last updated on 04/Sep/20

→r+1 =0 ⇒r=−1 ⇒y(x) =ae^(−x) +b y^′ =−ae^(−x) e⇒−ae^(−x) +ae^(−x) +b +7 =0 ⇒b=−7 ⇒ y(x) =ae^(−x) −7

$$\rightarrow\mathrm{r}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{r}=−\mathrm{1}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{ae}^{−\mathrm{x}} \:+\mathrm{b} \\ $$$$\mathrm{y}^{'} \:=−\mathrm{ae}^{−\mathrm{x}} \:\:\:\:\mathrm{e}\Rightarrow−\mathrm{ae}^{−\mathrm{x}} \:+\mathrm{ae}^{−\mathrm{x}} \:+\mathrm{b}\:+\mathrm{7}\:=\mathrm{0}\:\Rightarrow\mathrm{b}=−\mathrm{7}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{ae}^{−\mathrm{x}} −\mathrm{7} \\ $$

Answered by Aziztisffola last updated on 04/Sep/20

let y=uv ⇒y′=u′v+uv′ u′v+uv′+uv=−7 ⇔uv′+v(u′+u)=−7 u′+u=0 ⇒((u′)/u)=−1 ⇒∫((u′)/u)dx=∫−dx ln(u)=−x+C⇒u=e^(−x+C) =ke^(−x) ke^(−x) v′=−7⇒v′=−(7/k)e^x ⇒v=−(7/k)e^x +C y=ke^(−x) (−(7/k)e^x +C)=Ke^(−x) −7 (K∈R)

$$\:\mathrm{let}\:{y}={uv}\:\Rightarrow{y}'={u}'{v}+{uv}' \\ $$$$\:{u}'{v}+{uv}'+{uv}=−\mathrm{7}\:\Leftrightarrow{uv}'+{v}\left({u}'+{u}\right)=−\mathrm{7} \\ $$$$\:{u}'+{u}=\mathrm{0}\:\Rightarrow\frac{{u}'}{\mathrm{u}}=−\mathrm{1}\:\Rightarrow\int\frac{{u}'}{\mathrm{u}}\mathrm{dx}=\int−\mathrm{dx} \\ $$$$\:\mathrm{ln}\left({u}\right)=−{x}+{C}\Rightarrow{u}={e}^{−{x}+{C}} =\mathrm{k}{e}^{−{x}} \\ $$$$\:\mathrm{k}{e}^{−{x}} {v}'=−\mathrm{7}\Rightarrow{v}'=−\frac{\mathrm{7}}{\mathrm{k}}{e}^{{x}} \Rightarrow{v}=−\frac{\mathrm{7}}{\mathrm{k}}{e}^{{x}} +{C} \\ $$$${y}=\mathrm{k}{e}^{−{x}} \left(−\frac{\mathrm{7}}{\mathrm{k}}{e}^{{x}} +{C}\right)={Ke}^{−{x}} −\mathrm{7}\:\:\:\:\left({K}\in\mathbb{R}\right) \\ $$

Answered by john santu last updated on 04/Sep/20

HE: λ+1=0; λ=−1 ⇒y_(h ) = C.e^(−x) particular solution y_p = Ax+B comparing coeff y′=A, ⇒A+Ax+B+7 = 0 Ax + (B+A+7) = 0 { ((A=0)),((B=−7)) :} ⇒y_p = −7 general solution y_g = C.e^(−x) −7.

$${HE}:\:\lambda+\mathrm{1}=\mathrm{0};\:\lambda=−\mathrm{1}\:\Rightarrow{y}_{{h}\:} =\:{C}.{e}^{−{x}} \\ $$$${particular}\:{solution}\: \\ $$$${y}_{{p}} \:=\:{Ax}+{B}\: \\ $$$${comparing}\:{coeff}\: \\ $$$${y}'={A},\:\Rightarrow{A}+{Ax}+{B}+\mathrm{7}\:=\:\mathrm{0} \\ $$$${Ax}\:+\:\left({B}+{A}+\mathrm{7}\right)\:=\:\mathrm{0} \\ $$$$\:\begin{cases}{{A}=\mathrm{0}}\\{{B}=−\mathrm{7}}\end{cases}\:\Rightarrow{y}_{{p}} \:=\:−\mathrm{7}\: \\ $$$${general}\:{solution}\: \\ $$$${y}_{{g}} \:=\:{C}.{e}^{−{x}} −\mathrm{7}. \\ $$

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