## anonymous 5 years ago How do I get the solution for the differential equation 2y"-6y'+4y=7 ?

1. anonymous

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

2. anonymous

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.... $y(t)=C_1e^t+C_2e^{2t}+Y(t)\rightarrow y(t)=C_1e^t+C_2e^{2t}+\frac{7}{4}$

3. anonymous

So you integrated everything the first time right? And then you factored?

4. anonymous

I don't understand what you did to go from $2y" -> 2r^2 etc$

5. anonymous

I think I understand... I looked up a video

6. anonymous

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