# Match 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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To find the solution of the differential equation, we need to take the derivatives of the given functions and substitute them into the LS of the equation. If any of them add to equal zero, then that is the solution of the equation.

(1)

`y=x^{1/2}`

`y'=1/2x^{-1/2}`

`y''=-1/4x^{-3/2}`

Adding, we get

`LS=-1/4x^{-3/2}+7/2x^{-1/2}+12x^{1/2} ne 0`

This is not a solution.

(2)

`y=e^{-4x}`

`y'=-4e^{-4x}`

`y''=16e^{-4x}`

`LS=16e^{-4x}+7(-4e^{-4x})+12e^{-4x}=0=RS`

**This means that (2) is a solution to the differential equation.**

(3)

`y=sinx`

`y'=cos x`

`y''=-sinx`

`LS=-sinx+7cosx+12sinx ne 0`

This is not a solution.

(4)

`y=3x+x^2`

`y'=3+2x`

`y''=2`

`LS=2+7(3+2x)+12(3x+x^2)=12x^2+50x+23 ne0`

This is not a solution.

**The function (2) `y=e^{-4x}` is a solution.**