# How to solve the following initial value problem: y''+8y'-9=0, y(1)=1, y'(1)=0I think that the general solution is y=c1e^x + c2e^-9x but I could not get the particular solution. The books' answer...

How to solve the following initial value problem:

y''+8y'-9=0, y(1)=1, y'(1)=0

I think that the general solution is y=c1e^x + c2e^-9x but I could not

get the particular solution. The books' answer is

y=1/10e^(-9(t-1)) + 9/10e^(t-1).

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### 1 Answer

I think your equation is actually,

`y''+8y'-9y=0`

The solution is `y=c_1e^(t)+c_2e^(-9t)`

Because `a^2+8a-9 = (x-1)(x+9)` has solution 1 and -9

Now using y(1)=1 so substituting t=1 and y(1)=1 we get

`1=c_1e^(1)+c_2e^(-9(1))` solve for `c_2` to get

`c_2 = (1-c_1e^(1))/e^(-9) = e^9 - c_1e^10`

Taking the derivative of y we get

`y'=c_1e^(t)-9c_2e^(-9t)` and use y'(1)=0

`0=c_1e^1-9c_2e^(-9)` , so `c_1 = 9c_2e^(-9)/e^1 = 9c_2e^(-10)`

Substituting `c_2` from above we get

`c_1 = 9(e^9-c_1e^10)e^(-10) = 9e^-1 - 9c_1` this gives

`10c_1 = 9e^-1` so `c_1 = 9/10e^-1`

And `c_2 = c_1/9e^10 = 9/10e^-1/9e^10=1/10e^9`

So our answer is

` y=9/10e^-1e^t+1/10e^9e^(-9t)=9/10e^(t-1)+1/10e^(-9t+9)`

Which is equivalent to

` y=1/10e^(-9(t-1))+9/10e^(t-1)`