Particular Integral using Operator Method
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For a non-homogeneous differential equation with constant coefficients, $P(D)y=e^{mx}$, the particular integral $y_p$ is given by $$y_p= dfrac{e^{mx}}{P(m)}$$ if $P(m)neq 0 $ and if $m$ is a simple root of $P(m)$, then $P(m)=0$ , $$y_p=dfrac{x,e^{mx}}{G(m)} $$ and generally if $m$ is a root with multiplicity $r $ , $P(D)=(D-m)^rG(D)$ and $$ y_p=dfrac{x^r e^{mx}}{r! ;G(m)}$$ My qusetion is; what is the proof (intuition) behind this ?
calculus integration ordinary-differential-equations derivatives
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
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add a comment |
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For a non-homogeneous differential equation with constant coefficients, $P(D)y=e^{mx}$, the particular integral $y_p$ is given by $$y_p= dfrac{e^{mx}}{P(m)}$$ if $P(m)neq 0 $ and if $m$ is a simple root of $P(m)$, then $P(m)=0$ , $$y_p=dfrac{x,e^{mx}}{G(m)} $$ and generally if $m$ is a root with multiplicity $r $ , $P(D)=(D-m)^rG(D)$ and $$ y_p=dfrac{x^r e^{mx}}{r! ;G(m)}$$ My qusetion is; what is the proof (intuition) behind this ?
calculus integration ordinary-differential-equations derivatives
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
$endgroup$
add a comment |
$begingroup$
For a non-homogeneous differential equation with constant coefficients, $P(D)y=e^{mx}$, the particular integral $y_p$ is given by $$y_p= dfrac{e^{mx}}{P(m)}$$ if $P(m)neq 0 $ and if $m$ is a simple root of $P(m)$, then $P(m)=0$ , $$y_p=dfrac{x,e^{mx}}{G(m)} $$ and generally if $m$ is a root with multiplicity $r $ , $P(D)=(D-m)^rG(D)$ and $$ y_p=dfrac{x^r e^{mx}}{r! ;G(m)}$$ My qusetion is; what is the proof (intuition) behind this ?
calculus integration ordinary-differential-equations derivatives
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
$endgroup$
For a non-homogeneous differential equation with constant coefficients, $P(D)y=e^{mx}$, the particular integral $y_p$ is given by $$y_p= dfrac{e^{mx}}{P(m)}$$ if $P(m)neq 0 $ and if $m$ is a simple root of $P(m)$, then $P(m)=0$ , $$y_p=dfrac{x,e^{mx}}{G(m)} $$ and generally if $m$ is a root with multiplicity $r $ , $P(D)=(D-m)^rG(D)$ and $$ y_p=dfrac{x^r e^{mx}}{r! ;G(m)}$$ My qusetion is; what is the proof (intuition) behind this ?
calculus integration ordinary-differential-equations derivatives
calculus integration ordinary-differential-equations derivatives
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
asked Dec 13 '18 at 4:51
Aravindh VasuAravindh Vasu
13
13
add a comment |
add a comment |
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