How do I get the solution for the differential equation 2y"-6y'+4y=7 ?

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How do I get the solution for the differential equation 2y"-6y'+4y=7 ?

Mathematics
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What do you mean work out the problem? Thats what I'm trying to figure out. I think I would take the second integral from the first term, and the first integral from the second
I will use the method of undetermined coefficients: the characteristic equation is...\[2y''-6y'+4y=0\rightarrow 2r^2-6r+4=0\rightarrow (2r-2)(r-2)=0\] \[r_1=1, r_2=2\rightarrow y(t)=C_1e^t+C_2e^{2t}+Y(t)\] Now solve for Y(t) using the method of undetermined coefficients... \[Y(t)=A, Y'(t)=0, Y''(t)=0\] plug these values into the differential equation to determine Y(t)... \[Y(t)= 2(0)-6(0)+4(A)=7\rightarrow 4A=7\rightarrow Y(t)=A=\frac {7}{4}\] Thus.... \[y(t)=C_1e^t+C_2e^{2t}+Y(t)\rightarrow y(t)=C_1e^t+C_2e^{2t}+\frac{7}{4}\]
So you integrated everything the first time right? And then you factored?

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I don't understand what you did to go from \[2y" -> 2r^2 etc\]
I think I understand... I looked up a video
no.... i didn't integrate anything, this is a second order nonhomogenous differential equation so: \[ay''+by'+cy=g(t)\rightarrow ay''+by'+cy=0 , and, g(t)=0\] now think of a function that is its own derivative or a multiple of it.... \[y=e^{rt}\] is such a function, so...\[y=e^{rt}, y'=re^{rt}, y''=r^2e^{rt}\] now plug these in into the generic differential equation... \[ay''+by'+cy=g(t)\rightarrow ar^2e^{rt}+bre^{rt}+ce^{rt}=0\] now factor out the e^(rt) giving you the characteristic equation: \[(ar^2+br+c)e^{rt}\] since e^(rt) cant be 0, then ...\[ar^2+br+c=0\] and then find the roots of r

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