# Match each of the following differential equation with its solution: y''+ 7y'+ 12y = 0 1. y = x^(1/2) 2. y = e^(-4x) 3. y = sin(x) 4. y = 3x + x^(2)

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### 1 Answer

You should substitute `e^(lambda x)` for y in equation above such that:

`(d^2(e^(lambda x)))/(dx^2)+ 7(d(e^(lambda x)))/(dx) + 12e^(lambda x) = 0`

`lambda^2*e^(lambda x) + 7lambda*e^(lambda x) + 12e^(lambda x) = 0`

You need to factor out `e^(lambda x)` such that:

`e^(lambda x)(lambda^2 + 7lambda + 12) = 0`

Since `e^(lambda x)!=0` , hence you should solve for lambda the equation `lambda^2 + 7lambda + 12 = 0 ` such that:

`lambda^2 + 7lambda + 12 = 0`

`lambda_(1,2) = (-7+-sqrt(49 - 48))/2`

`lambda_(1,2) = (-7+-1)/2 => lambda_1 = -3`

`lambda_2 = -4`

If `lambda_1 = -3 =>y_1(x) = c_1*e^(-3x).`

If `lambda_1 = -4 => y_2(x) = c_2*e^(-4x).`

**Hence, the evaluating the general solution to the given equation yields `y(x) = c_1*e^(-3x) + c_2*e^(-4x)` , hence, the second option `2. y = e^(-4x),` represents one of the solutions to the give ODE.**

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