y ′ ′ − y ′ − 12 y = t e 4 t λ 2 − λ − 12 = 0 ⇒ λ ∈ { 4 , − 3 } T h u s t h e f o r m o f p a r t i c u l a r s o l u t i o n i s t P ( t ) e 4 t , P i s a p o l y n o m i a l o f p o w e r 1 y = t ( A + B t ) e 4 t = ( A t + B t 2 ) e 4 t y ′ = ( A + ( 2 B + 4 A ) t + 4 B t 2 ) e 4 t y ′ ′ = ( 8 A + 2 B + ( 16 A + 16 B ) t + 16 B t 2 ) e 4 t ( 8 A + 2 B + ( 16 A + 16 B ) t + 16 B t 2 − A − ( 2 B + 4 A ) t − 4 B t 2 − 12 A t − 12 B t 2 ) e 4 t = t e 4 t { 7 A + 2 B = 0 14 B = 1 0 B = 0 ⇒ { A = − 1 49 B = 1 14 y = ( − 1 49 t + 1 14 t 2 ) e 4 t y''-y'-12y=te^{4t}\\\lambda ^2-\lambda -12=0\Rightarrow \lambda \in \left\{ 4,-3 \right\} \\Thus\,\,the\,\,form\,\,of\,\,particular\,\,solution\,\,is\,\,tP\left( t \right) e^{4t},P\,\,is\,\,a\,\,polynomial\,\,of\,\,power\,\,1\\y=t\left( A+Bt \right) e^{4t}=\left( At+Bt^2 \right) e^{4t}\\y'=\left( A+\left( 2B+4A \right) t+4Bt^2 \right) e^{4t}\\y''=\left( 8A+2B+\left( 16A+16B \right) t+16Bt^2 \right) e^{4t}\\\left( 8A+2B+\left( 16A+16B \right) t+16Bt^2-A-\left( 2B+4A \right) t-4Bt^2-12At-12Bt^2 \right) e^{4t}=te^{4t}\\\left\{ \begin{array}{c} 7A+2B=0\\ 14B=1\\ 0B=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} A=-\frac{1}{49}\\ B=\frac{1}{14}\\\end{array} \right. \\y=\left( -\frac{1}{49}t+\frac{1}{14}t^2 \right) e^{4t}\\ y ′′ − y ′ − 12 y = t e 4 t λ 2 − λ − 12 = 0 ⇒ λ ∈ { 4 , − 3 } T h u s t h e f or m o f p a r t i c u l a r so l u t i o n i s tP ( t ) e 4 t , P i s a p o l y n o mia l o f p o w er 1 y = t ( A + Bt ) e 4 t = ( A t + B t 2 ) e 4 t y ′ = ( A + ( 2 B + 4 A ) t + 4 B t 2 ) e 4 t y ′′ = ( 8 A + 2 B + ( 16 A + 16 B ) t + 16 B t 2 ) e 4 t ( 8 A + 2 B + ( 16 A + 16 B ) t + 16 B t 2 − A − ( 2 B + 4 A ) t − 4 B t 2 − 12 A t − 12 B t 2 ) e 4 t = t e 4 t ⎩ ⎨ ⎧ 7 A + 2 B = 0 14 B = 1 0 B = 0 ⇒ { A = − 49 1 B = 14 1 y = ( − 49 1 t + 14 1 t 2 ) e 4 t
#339914 on Dec 2023
Comments