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

### 1 Answer | Add Yours

When you get the derivative of the above functions; function 2 and 3 goes with same manner since they have `e^x` terms and sinx terms. In its derivative no x terms will occur.

eg:

y = `e^(-4x)` y = sinx

y' = `-4e^(-4x)` y' = cosx

Here we can see There is no individual x terms in derivative. If you get the second derivative it will remain same.

`2(x^2)y''+ 3xy" = y`

For this to happen we need x terms in the derivatives.

So the answer will come from either function 1 or 4.

`y = x^(1/2)`

`y' = 1/(2sqrtx) = (1/2)(x^(-1/2))`

`y'' = (1/2)(-1/2)x^(-3/2) = (-1/4)x^(-3/2)`

`2(x^2)y''+ 3xy"`

`= 2(x^2)(-1/4)x^(-3/2)+3x(-1/4)x^(-3/2)`

`= (-1/2)x^(1/2)+(-3/4)x^(-1/2)`

`y = 3x + x^(2)`

`y' = 3+2x`

`y'' = 2`

`2(x^2)y''+ 3xy"`

`= 2x^2*2+3x*2`

`= 4x^2+6x`

**So non of the answers satisfies the required criteria.**

**Sources:**