Substituting the exponential trial solution into the homogeneous equation (Q1662): Difference between revisions

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y=e^{rx},\\quad y'=r\\,e^{rx},\\quad y''=r^{2}\\,e^{rx};\\qquad y''+a\\,y'+b\\,y=\\bigl(r^{2}+a\\,r+b\\bigr)e^{rx}=0\\ \\Rightarrow\\ r^{2}+a\\,r+b=0,\\quad e^{rx}\\neq 0
 
Property / LaTeX source: y=e^{rx},\\quad y'=r\\,e^{rx},\\quad y''=r^{2}\\,e^{rx};\\qquad y''+a\\,y'+b\\,y=\\bigl(r^{2}+a\\,r+b\\bigr)e^{rx}=0\\ \\Rightarrow\\ r^{2}+a\\,r+b=0,\\quad e^{rx}\\neq 0 / rank
Normal rank
 
Property / instance of
 
Property / instance of: mathematical expression / rank
 
Normal rank
Property / LaTeX source
 
y=e^{rx},\\quad y'=r\\,e^{rx},\\quad y''=r^{2}\\,e^{rx};\\qquad y''+a\\,y'+b\\,y=\\bigl(r^{2}+a\\,r+b\\bigr)e^{rx}=0\\ \\Rightarrow\\ r^{2}+a\\,r+b=0,\\quad e^{rx}\\neq 0
Property / LaTeX source: y=e^{rx},\\quad y'=r\\,e^{rx},\\quad y''=r^{2}\\,e^{rx};\\qquad y''+a\\,y'+b\\,y=\\bigl(r^{2}+a\\,r+b\\bigr)e^{rx}=0\\ \\Rightarrow\\ r^{2}+a\\,r+b=0,\\quad e^{rx}\\neq 0 / rank
 
Normal rank

Latest revision as of 07:28, 4 September 2026

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Substituting the exponential trial solution into the homogeneous equation
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    y=e^{rx},\\quad y'=r\\,e^{rx},\\quad y''=r^{2}\\,e^{rx};\\qquad y''+a\\,y'+b\\,y=\\bigl(r^{2}+a\\,r+b\\bigr)e^{rx}=0\\ \\Rightarrow\\ r^{2}+a\\,r+b=0,\\quad e^{rx}\\neq 0
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