# Verify that the functions `y_1(x)=e^(-2x)` , `y2(x)=e^(-3x)` `y'' + 5y' + 6y = 0` and find the solution `y(0)=1` , `y'(0)=1`

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The problem lets you off too easily (I would have made you find `y_1` and `y_2`)! So, we're going to verify that those two functions are solutions to the second-order DE:

Checking `y_1`

`y_1' = -2e^(-2x)` `y_1'' = 4e^(-2x)`

`y_1''+5y_1' +6y_1 = 4e^(-2x) + 5*(-2e^(-2x)) + 6e^(-2x) = 4-10+6 = 0`

Looks like `y_1` is a solution.

Now for `y_2`:

`y_2' = -3e^(-3x)` `y_2'' = 9e^(-3x)`

`y_2'' + 5y_2' + 6y_2 = 9e^(-3x) + 5(-3e^(-3x)) +6e^(-3x) = 9-15+6 = 0`

Looks like `y_2` is also a solution!

Now, we need to find the overall solution to the equation given the initial conditions: `y(0) = 1, y'(0) = 1`

Well, we need to start off by reconizing something interesting about the functions `y_1` and `y_2`. If you plug in any linear combination of the two functions, you end up with another solution to the differential equation. In other words, given:

`y = A_1y_1+A_2y_2=A_1e^(-2x) + A_2e^(-3x)`

So, we're going to let the following function be y, and we show the derivative of y, y':

`y = A_1e^(-2x) + A_2e^(-3x)`

`y' = -2A_2e^(-2x) -3A_2e^(-3x)`

you could show in the same way we did above that `y` is a solution to the D.E.

So, we need to find `y` given those initial conditions, meaning we solve for the constants `A_1 and A_2`.

Let's plug the values in to see what we get!

`1 = A_1e^(-2*0) + A_2e^(-3*0)`

`1 = -2A_1e^(-2*0) - 3 A_2e^(-3*0)`

To solve this, let's just quickly recognize that `e^0 = 1`. Now, we just have a simple system of equations:

`1 = A_1 + A_2`

`1 = -2A_1-3A_2`

Now, you could solve this with a ton of methods. However, I'm just going to skip this tedious Algebra I calculation and give you the result:

`A_1 = 4`

`A_2 = -3`

If you just HAVE to know, I just found the reduced-row echelon form of the augmented matrix of coefficients. It was 2 steps!

Now, we have our final equation for y:

`y = 4e^(-2x) - 3e^(-3x)`

Hope that helps!

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