# Match y = sin(x) 2. y = e^(-4x) differential equation with its solution: a.y''+y = 0 b.y'' + 7y' + 12y = 0

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### 1 Answer

You need to solve the second order linear ODE `y'' + 7y' + 12y = 0,` hence, you should consider `y(x) = e^(lambda x)` .

Substituting e^(lambda x) for y(x) into the equation yields:

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

You should come up with the following substitution such that:

`(d^2(e^(lambda x)))/(dx^2) = lambda^2*e^(lambda x)`

`(d(e^(lambda x)))/(dx) = lambda*e^(lambda x)`

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

Factoring out `e^(lambda x) ` yields:

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

Since `e^(lambda x)!=0` , then `lambda^2 + 7lambda + 12 = 0` .

You need to 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`

Hence , if `lambda = -3 => y_1(x) = c_1e^(-3x)` and if `lambda = -4 => y_2(x) = c_2e^(-4x).`

**Hence, evaluating the general solution to the given second order linear ODE `y'' + 7y' + 12y = 0` yields`y(x) = c_1e^(-3x) + c_2e^(-4x)` , hence, the second option matches the best to the solution to this equation.**

You need to solve the second order linear ODE `y''+y = 0` , hence,you should consider `y(x) = e^(lambda x). `

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

Factoring out `e^(lambda x)` yields:

`e^(lambda x)(lambda^2 + 1) = 0`

Since `e^(lambda x)!=0` , then `lambda^2 +1 = 0 ` such that:

`lambda^2 +1 = 0`

`lambda_1 =i ; lambda_2 = -i`

Hence , if `lambda =i => y_1(x) = c_1e^(ix)` and if `lambda = -i => y_2(x) = c_2e^(-ix)` .

`y(x) = c_1cos x + c_2sin x`

**Hence, evaluating the general solution to the given second order linear ODE `y''+y = 0` yields `y(x) = c_1cos x + c_2sin x` , hence, the first option matches the best to the solution to this equation.**

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