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anonymous
 5 years ago
how do i solve y'' +2y + y = 6 and y(0) and y'(0)=2
anonymous
 5 years ago
how do i solve y'' +2y + y = 6 and y(0) and y'(0)=2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You would notice the equation is the second derivative. y'(0)=2 is first derivative. So you have to work backwards. Integrate. You have to do a bit of reading of your text book to learn how to find C.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the 2nd term 2y or 2y'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use undetermined coefficeints or variation of parameters since this is non homogenous. You want the general solution which is yp+yc. Solve for the characteristic equation first so yc, which is a quadratic equation looking at the coefficients of y'', y' and y. So you have r^2+2+1=0 So you get (r+1)^2. You have 1 as a repeated root. So yc=c1+e^t+c2te^t. Now to solve yp, you have to make a guess. Since g(t)=6, your guess is yp=A. Now normally you differentiate this twice and plug back into original D.E and try to solve for the constant. You technically are still doing this, but it is quite obvious you will get zero and A=6. Therefore, yp=6. So, as we said in the beginning your general solution is yp+yc. So add those two up, and that is your solution. Only thing left to do is to apply your initial conditions and solve for c1 and c2. So you need to differentiate this general solution to apply the second initial condition to solve for the constants then you're done.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this is a non homogeneous differential equation with constant coefficients. the general solution of such an equation would be in the form: \[y(x)=y_c(x)+y_p(x)\] where y_c is the solution associated with the its homogeneous equation, and y_p is the particular solution.
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