been working away on practice problems for my final exam. I have a worksheet that has a bunch of differential equations with an initial value problem followed by some general solution ones. Ive been on a break and can't seem to shake off the proverbial "rust." I just need help on direction because I know the answer
I need to find the general solution of:
y''(x)+y'(x)-2y(x)=f(x) where f(x)=e^(-x)
I know the answer is supposed to be:
y(x) = c_1 e^(-2 x)+c_2 e^x-e^(-x)/2
do I start by rewriting the function as r^2+r-2=0?
Also, I already solved the initial value problem just before this one (same diff Eq with y(0)=0 and y'(0)=1 when f(x)=0). is any of the stuff I got in that one of use for this one or do I have to start from the beginning?
From the previous equation I have:
y=C_1_e^(x) + C_2_e^(-2x)
y"=C_1_e^(x) - 2C_2_e^(-2x)
as well as the final answer of the initial value problem being: y=(1/3)e^x - (1/3)e^(-2x)
Any help on direction is much appreciated. when it comes to math (math's in the UK where I am studying), figuring out a direction is what seems to get me. Thanks!
I need to find the general solution of:
y''(x)+y'(x)-2y(x)=f(x) where f(x)=e^(-x)
I know the answer is supposed to be:
y(x) = c_1 e^(-2 x)+c_2 e^x-e^(-x)/2
do I start by rewriting the function as r^2+r-2=0?
Also, I already solved the initial value problem just before this one (same diff Eq with y(0)=0 and y'(0)=1 when f(x)=0). is any of the stuff I got in that one of use for this one or do I have to start from the beginning?
From the previous equation I have:
y=C_1_e^(x) + C_2_e^(-2x)
y"=C_1_e^(x) - 2C_2_e^(-2x)
as well as the final answer of the initial value problem being: y=(1/3)e^x - (1/3)e^(-2x)
Any help on direction is much appreciated. when it comes to math (math's in the UK where I am studying), figuring out a direction is what seems to get me. Thanks!