creative_22 :)
New member
The question is: Show that if lambda denotes a positive constant, and b is any real number, then any solution, say phi, of the ODE y'(x)+y(x)=be^-lambdax, has the property that phi->0 as x->+infinity.
Comment: Be sure to treat all possible cases. To start, find the general solution of the given equation.
For the homogeneous solution I got y(h)=c1e^-x and for the particular solution I used the equations
y(p)= Abe^-lambdax + Bce^lambdax and y'(p)=-lambdabAe^-lambdax+lambdacBe^lambdax
Putting y and y' into the ODE to get B and c =0 and A=1/(-lambda+1). This makes the particular solution y(p)= b/(-lambda+1)*e^(-lambdax).
Thus this makes the general solution to the ODE y=c1e^-x+b/(-lambda+1)*e^-lamdax
My problem is that I am not sure what it means to test all possible cases and I am having trouble figuring out how to determine what phi and x go to. Also, my general solution could be a little off. Any advice or help that you can give me would be great. Thank you.
Comment: Be sure to treat all possible cases. To start, find the general solution of the given equation.
For the homogeneous solution I got y(h)=c1e^-x and for the particular solution I used the equations
y(p)= Abe^-lambdax + Bce^lambdax and y'(p)=-lambdabAe^-lambdax+lambdacBe^lambdax
Putting y and y' into the ODE to get B and c =0 and A=1/(-lambda+1). This makes the particular solution y(p)= b/(-lambda+1)*e^(-lambdax).
Thus this makes the general solution to the ODE y=c1e^-x+b/(-lambda+1)*e^-lamdax
My problem is that I am not sure what it means to test all possible cases and I am having trouble figuring out how to determine what phi and x go to. Also, my general solution could be a little off. Any advice or help that you can give me would be great. Thank you.