{ (((d^2 y/dx^2 )−2 (dy/dx)+y =3e^(4x) )),((y(0)=−(2/3) ; (dy/dx)∣


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Question Number 157141 by cortano last updated on 20/Oct/21
${ (((d^2 y/dx^2 )−2 (dy/dx)+y =3e^(4x) )),((y(0)=−(2/3) ; (dy/dx)∣_(x=0) = ((13)/3))) :}$
${\begin{cases} \frac{d^{2} y}{{d x}^{2}} - 2 \frac{d y}{d x} + y = 3 e^{4 x} \\ y (0) = - \frac{2}{3}; \frac{d y}{d x} ∣_{x = 0} = \frac{13}{3} \end{cases}$
Commented by john_santu last updated on 21/Oct/21
${ ((y′′−2y′+y=3e^(4x) )),((y(0)=−(2/3) ; y′(0)=((13)/3))) :} By wronskian method ⇒λ^2 −2λ+1=0⇒λ_(1,2) =1 ⇒y_h = Ae^x +Bxe^x Particular solution W(u_1 ,u_2 )= determinant (((e^x xe^x )),((e^x e^x +xe^x )))=e^(2x) W_1 = determinant (((0 xe^x )),((3e^(4x) e^x +xe^x )))=−3xe^(5x) W_2 = determinant (((e^x 0)),((e^x 3e^(4x) )))=3e^(5x) ⇒v_1 =∫ (w_1 /w) dx=∫ ((−3xe^(5x) )/e^(2x) ) dx =−3∫xe^(3x) dx =−3[ (1/3)xe^(3x) −(1/9)e^(3x) ]=−xe^(3x) +(1/3)e^(3x) ⇒v_2 =∫ (w_2 /w) dx=∫ ((3e^(5x) )/e^(2x) ) dx=∫3e^(3x) dx=e^(3x) ⇒y_p =u_1 v_1 +u_2 v_2 =e^x (−xe^(3x) +(1/3)e^(3x) )+xe^x (e^(3x) ) ⇒y_p =(1/3)e^(4x) y_g =Ae^x +Bxe^x +(1/3)e^(4x) { ((x=0⇒A+(1/3)=−(2/3) ; A=−1)),((y′=Ae^x +Be^x +Bxe^x +(4/3)e^(4x) )),((y′(0)=−1+B+(4/3)=((13)/3)⇒B=4 )) :} ⇔ ∴ y=−e^x +4xe^x +(1/3)e^(4x)$
${\begin{cases} y^{″} - 2 y^{'} + y = 3 e^{4 x} \\ y (0) = - \frac{2}{3}; y^{'} (0) = \frac{13}{3} \end{cases}$ $B y w r o n s k i a n m e t h o d$ $\Rightarrow λ^{2} - 2 λ + 1 = 0 \Rightarrow λ_{1, 2} = 1$ $\Rightarrow y_{h} = {A e}^{x} + {B x e}^{x}$ $P a r t i c u l a r s o l u t i o n$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{x} {x e}^{x} \\ e^{x} e^{x} + {x e}^{x} \end{matrix} \| = e^{2 x}$ $W_{1} = \| \begin{matrix} 0 {x e}^{x} \\ 3 e^{4 x} e^{x} + {x e}^{x} \end{matrix} \| = - 3 {x e}^{5 x}$ $W_{2} = \| \begin{matrix} e^{x} 0 \\ e^{x} 3 e^{4 x} \end{matrix} \| = 3 e^{5 x}$ $\Rightarrow v_{1} = \int \frac{w_{1}}{w} d x = \int \frac{- 3 {x e}^{5 x}}{e^{2 x}} d x$ $= - 3 \int {x e}^{3 x} d x = - 3 [\frac{1}{3} {x e}^{3 x} - \frac{1}{9} e^{3 x}] = - {x e}^{3 x} + \frac{1}{3} e^{3 x}$ $\Rightarrow v_{2} = \int \frac{w_{2}}{w} d x = \int \frac{3 e^{5 x}}{e^{2 x}} d x = \int 3 e^{3 x} d x = e^{3 x}$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{x} (- {x e}^{3 x} + \frac{1}{3} e^{3 x}) + {x e}^{x} (e^{3 x})$ $\Rightarrow y_{p} = \frac{1}{3} e^{4 x}$ $y_{g} = {A e}^{x} + {B x e}^{x} + \frac{1}{3} e^{4 x}$ ${\begin{cases} x = 0 \Rightarrow A + \frac{1}{3} = - \frac{2}{3}; A = - 1 \\ y^{'} = {A e}^{x} + {B e}^{x} + {B x e}^{x} + \frac{4}{3} e^{4 x} \\ y^{'} (0) = - 1 + B + \frac{4}{3} = \frac{13}{3} \Rightarrow B = 4 \end{cases}$ $\Leftrightarrow ∴ y = - e^{x} + 4 {x e}^{x} + \frac{1}{3} e^{4 x}$
	Answered by qaz last updated on 21/Oct/21
	$y_p =(1/(D^2 −2D+1))3e^(4x) =(1/(4^2 −2∙4+1))3e^(4x) =(1/3)e^(4x) ⇒y=(C_1 +C_2 x)e^x +(1/3)e^(4x) { ((y(0)=C_1 +(1/3)=−(2/3))),((y(0)′=C_2 +C_1 +(4/3)=((13)/3))) :} ⇒C_1 =−1 C_2 =4 ⇒y=(4x−1)e^x +(1/3)e^(4x)$
	$y_{p} = \frac{1}{D^{2} - 2 D + 1} {3 e}^{4 x} = \frac{1}{4^{2} - 2 \cdot 4 + 1} {3 e}^{4 x} = \frac{1}{3} e^{4 x}$ $\Rightarrow y = (C_{1} + C_{2} x) e^{x} + \frac{1}{3} e^{4 x}$ ${\begin{cases} y (0) = C_{1} + \frac{1}{3} = - \frac{2}{3} \\ y {(0)}^{'} = C_{2} + C_{1} + \frac{4}{3} = \frac{13}{3} \end{cases}$ $\Rightarrow C_{1} = - 1 C_{2} = 4$ $\Rightarrow y = (4 x - 1) e^{x} + \frac{1}{3} e^{4 x}$
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