Second-Order Homogeneous Linear Equation

[andrewdavid]

I have this intial value problem: y''-4y'-5y=0, y(1)=0, y'(1)=2. My AUX equation is r^2-4r-5=0. I factor and get r=5, r=-1 and my equation becomes y(x)=C1e^(5x)+C2e^(-1x) (C1 and C2 are constants). I took the derivative of y(x) and then tried to use my initial value's to solve for C1 and C2. I got C1e^5+C2e^(-1)=0 and 5C1e^5-C2e^(-1)=2. For some reason when I solve for C1 or C2 and put them back in the equations it doesn't check out. I'd appreciate any advice to help me find what C1 and C2 are. Thanks!

 

[Data]

C_1e^5 + C_2e^{-1} = 0, \ 5C_1e^5 - C_2e^{-1} = 2 \Longrightarrow 6C_1e^5 = 2 \Longrightarrow C_1 = \frac{1}{3}e^{-5}

so

C_1e^5 + C_2e^{-1} = \frac{1}{3}+C_2e^{-1}=0 \Longrightarrow C_2 = -\frac{e}{3}.

 

[M98Ranger]

It looks like you are on the right track with your solution for the second-order homogeneous linear equation. However, when you took the derivative of y(x), you made a mistake. The derivative of C1e^(5x) is not C1e^5, it is 5C1e^(5x). Similarly, the derivative of C2e^(-1x) is not -C2e^(-1), it is -C2e^(-1x). This error is causing your initial value equations to not check out.

To solve for C1 and C2, you can use the initial value equations as follows:

y(1) = C1e^(5*1) + C2e^(-1*1) = C1e^5 + C2e^(-1) = 0
y'(1) = 5C1e^(5*1) - C2e^(-1*1) = 5C1e^5 - C2e^(-1) = 2

Now you have two equations with two unknowns (C1 and C2). You can solve this system of equations using substitution or elimination to find the values of C1 and C2. Once you have those values, you can plug them back into your original equation to check if they satisfy the initial value problem.

I hope this helps and good luck with your problem!