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Calculus and Beyond Homework Help
X' = ax+b find general solution
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[QUOTE="HallsofIvy, post: 5183229, member: 637751"] There are a number of different ways to do a problem like that. Probably the simplest is: Letting [tex]x= \begin{pmatrix}x_1(t) \\ x_2(t)\end{pmatrix}[/tex] we can write the equation as [tex]\begin{pmatrix}x_1' \\ x_2'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -t^{-2} & t^{-1}\end{pmatrix}\begin{pmatrix}x_1\\ x_2\end{pmatrix}+ \begin{pmatrix}0 \\ 2t^{-1}\end{pmatrix}[/tex] [tex]\begin{pmatrix}x_1' \\ x_2'\end{pmatrix}= \begin{pmatrix}x_2 \\ -t^{-2}x_1+ t^{-1}x_2+ 2t^{-1}\end{pmatrix}[/tex] which gives the two equations [itex]x_1'= x_2[/itex] and [itex]x_2'= -t^{-2}x_1+t^{-1}x_2+ 2t^{-1}[/itex]. To solve those equations, differentiate the first one a second time: [itex]x_1''= x_2'= -t^{-2}x_1+ t^{-1}x_2+ 2t^{-1}[/itex]. From the first of the two equations, [itex]x_2= x_1'[/itex] so we can write this equation as [itex]x_1''= -t^{-2}x_1+ t^{-1}x_1'+ 2t^{-1}[/itex] which is the single second order equation [itex]x_1''- t^{-1}x_1'+ t^{-2}x_1= 2t^{-1}[/itex]. Multiply by [itex]t^2[/itex] to get [itex]t^2x_1''- tx_1'+ x_1= 2t[/itex]. That is a "Cauchy type" or "equi-potential" equation (since each derivative is multiplied by t to the same power as the order of the derivative). The change of variable, u= ln(t), changes the equation to a corresponding "constant coefficients" equation, [itex]d^2x_1/du^2- 2 dx_1/du+ x_1= 2e^u[/itex]. Can you solve that equation? [/QUOTE]
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X' = ax+b find general solution
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