Question

Question #31289

Answer 1

`y = 2/3 x^3 +x^2-2/3 x+ 2`

Explanation:

If `(d^2y)/dx^2 = 2-4x` then `(dy)/(dx)=2x-1/2xx 4 x^2+ a` and

`y = 1/2 xx 2 x^2-1/3 1/2 xx 4 x^3 + a x + b = x^2+2/3 x^3+a x+ b`

Now we know

`{((-1)^2+2/3(-1)^3+a(-1)+b = 3),(0^2+2/3 0^3+a cdot 0+b = 2):}`

Solving for `a,b` we have

`a=-2/3, b = 2` so the curve is

`y = 2/3 x^3 +x^2-2/3 x+ 2`

Answer 2

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

Explanation:

This is a separable differential equation, so we can find the general solution by integration:

`(d^2y)/(dx^2) = 2-4x`

`(dy)/(dx) = int (2-4x)dx = 2x-2x^2+C_1`

`y = int (2x-2x^2+C_1)dx = x^2-2/3x^3+C_1x+C_2`

We can now find the values of `C_1` and `C_2` from the equations we get from the initial conditions:

`y(-1) = 3`

`y(0) = 2`

so that:

`{(1+2/3-C_1+C_2 = 3),(C_2 = 2):}`

`{(C_1 = -3+2+1+2/3 = 2/3),(C_2 = 2):}`

and we can conclude that:

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