Solve the boundary-value problem.

y''+5y'-6y=0, y(0)=0, y(2)=1
r^2+5r-6=0, r1=1, r2=-6
y=c1*e^x + c2*e^-6x
y(0)=c1+c2=0, c1=-c2

Is this correct? Is there a way to further simplify this for y(x)? Thanks.

No. The second term has 6e^-6x in it, I don't see where the coefficent six came from. Testing it to the initial conditions, at x=0, y does not equal zero with that six in.

Oh, thank you for catching my mistake. I got the correct answer now. Thanks!

You're welcome! I'm glad I was able to catch the mistake. It's important to double-check the calculations to ensure accuracy.

Now let's simplify the equation further for y(x). Recall the expression we obtained:

y(x) = (e^x) / (e^2 - 6e^(-12)) + (6e^(-6x)) / (e^2 - 6e^(-12))

To simplify further, notice that e^(-12) is a very small number and can be considered negligible compared to e^2. Therefore, we can approximate the denominator (e^2 - 6e^(-12)) as just e^2.

Now our simplified solution becomes:

y(x) ≈ (e^x) / e^2 + (6e^(-6x)) / e^2

By dividing the terms in the solution by e^2, we can rewrite it as:

y(x) ≈ (1 / e^2) * e^x + (6 / e^2) * e^(-6x)

Further simplification can be done by recognizing that (1 / e^2) and (6 / e^2) are just constants. We can replace them by a and b, respectively:

y(x) ≈ a * e^x + b * e^(-6x)

So in the end, the simplified form of the solution is:

y(x) ≈ (1 / e^2) * e^x + (6 / e^2) * e^(-6x)

where a = 1 / e^2 and b = 6 / e^2.

I apologize for the oversight in the previous response. Thank you for catching it, and I hope this clarification helps!