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 January 31st, 2012, 03:51 PM #1 Newbie   Joined: Jan 2012 Posts: 25 Thanks: 0 Ordinary differential equation y' = 2y + 4 Hey guys, I have to solve this by finding a constant solution and writing y(t) = y_h(t) + y_p(t). So far I have: y' = 2y + 4 y_p = at + b (at + b)' - 2(at + b) = 4 a - 2at - 2b = 4. Now I am stuck. Also, my notes say to set y' = 0, but I am not sure what that means exactly. Any help?
 January 31st, 2012, 05:23 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Ordinary differential equation y' = 2y + 4 We are given the ODE: $y'(t)=2y(t)+4$ If I were after the general solution, I would first arrange it as: $y'(t)-2y(t)=4$ then multiply through by the integrating factor $e^{-2t}$: $e^{-2t}y'(t)-2e^{-2t}y(t)=4e^{-2t}$ $e^{-2t}y'(t)+$$-2e^{-2t}$$y(t)=4e^{-2t}$ $\frac{d}{dt}$$e^{-2t}y$$=4e^{-2t}$ Integrate with respect to t: $\int\frac{d}{dt}$$e^{-2t}y$$\,dt=-2\int e^{-2t}\,$$-2\,dt$$$ $e^{-2t}y=-2e^{-2t}+C$ $y(t)=Ce^{2t}-2$ However, you are instructed to find the solution to the corresponding homogeneous equation and a particular solution: $y'(t)-2y(t)=0$ The associated auxiliary equation is: $r-2=0$ Thus: $y_h(t)=c_1e^{2t}$ Using the method of undetermined coefficients, we assume a particular solution of the form: $y_p(t)=A$ where A is an arbitrary real constant, and thus: $y_p'(t)=0$ Substituting into the ODE gives: $0-2A=4$ $A=-2$ Thus, by the principle of superposition, we have: $y(t)=y_h(t)+y_p(t)=c_1e^{2t}-2$
January 31st, 2012, 11:57 PM   #3
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Re: Ordinary differential equation y' = 2y + 4

Quote:
 Originally Posted by Norm850 Hey guys, I have to solve this by finding a constant solution and writing y(t) = y_h(t) + y_p(t). So far I have: y' = 2y + 4 y_p = at + b (at + b)' - 2(at + b) = 4 a - 2at - 2b = 4. Now I am stuck. Also, my notes say to set y' = 0, but I am not sure what that means exactly. Any help?
a - 2at - 2b = 4. any value of t. So a = 0 and -2b=4
But this gives ony one solution of the ODE, not all.

dy/dx=2y+4 = 2(y+2)
dy/(y+2) ) =2 dx
ln(y+2)=2x+constant
y=C*exp(2x) -2

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