How to solve the following initial value problem:
`y''+8y'-9y=0, y(1)=1, y'(1)=0`
I think that the general solution is `y=c_1e^x + c_2e^-9x` but I could not
get the particular solution. The books' answer is
`y=1/10e^(-9(t-1)) + 9/10e^(t-1)` .
1 Answer | Add Yours
Alright, so I'll start by pointing out that this is a homogeneous second-order differential equation. Because it is homogeneous, I just wanted to point out that there is no particular solution! Don't worry about that!
Let's find the solution:
`y'' + 8y' -9y=0`
We notice that the equation is a D.E. with constant coefficients, and recall that the solution will be of the form:
` ` `y = c_1e^(lambda_1t) + c_2 e^(lambda_2t)`
where `c_1` and `c_2` are constants of integration (solved-for by using the initial conditions) and `lambda_1` and `lambda_2` being the solution to the characteristic equation.
We derive the characteristic equation by pointing out that, based on our basis function (`e^(lambdat)`) and the differential equation above that our solution is found by solving:
`lambda^2 + 8lambda - 9 = 0`
This, when solved-for, gives you `lambda = -9, 1`, So, our equation becomes exactly what you had!
`y = c_1e^x + c_2e^(-9x)`
Now, we solve for the initial conditions. We know `y(1) = 1` and `y'(1) = 0`.
We need to figure out what y' is to determine the constants:
`y' = c_1e^x -9c_2e^-9x`
So, with our initial values, we get two equations:
`1 = c_1e + c_2e^(-9)`
`0 =c_1e -9c_2e^-9`
Using the second equation, we can find an easily-substitutable expression for `c_1e`:
`c_1e = 9c_2e^-9`
So, let's substitute this into the first equation:
`1 = 9c_2e^-9 + c_2e^-9`
`1 = 10c_2e^-9`
Dividing by `10e^-9` we get our value for `c_2`:
`c_2 = 1/10 * e^9`
Now, let's solve for `c_1` by using the equation we used to substitute before:
`c_1e = 9c_2e^-9`
`c_1e = 9(1/10e^9)e^-9`
Notice that the e terms on the rigth cancel, giving us:
`c_1e = 9/10`
Dividing by e (which is the same as multiplying by `e^-1`):
`c_1 = 9/10 e^-1`
So, we have our values for `c_1` and `c_2`, and now we can find our full equation:
`y = 9/10e^-1e^x + 1/10e^9e^(-9x)`
Now, we can combine the exponents (remember `e^ae^b = e^(a+b)`)
`y = 9/10 e^(x-1) + 1/10 e^(-9x+9)`
Now, we factor out the 9 in the exponent of the right-sided term:
`y = 9/10e^(x-1) + 1/10 e^(-9(x-1))`
Now, we (just for fun times) let t = x:
`y = 9/10 e^(t-1) + 1/10e^(-9(t-1))`
And there's your answer! You had it totally correct! You just needed to stick with the problem, and you would have gotten it!
Hope that helps!
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