# Question #31289

Mar 29, 2017

$y = \frac{2}{3} {x}^{3} + {x}^{2} - \frac{2}{3} x + 2$

#### Explanation:

If $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 - 4 x$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - \frac{1}{2} \times 4 {x}^{2} + a$ and

$y = \frac{1}{2} \times 2 {x}^{2} - \frac{1}{3} \frac{1}{2} \times 4 {x}^{3} + a x + b = {x}^{2} + \frac{2}{3} {x}^{3} + a x + b$

Now we know

$\left\{\begin{matrix}{\left(- 1\right)}^{2} + \frac{2}{3} {\left(- 1\right)}^{3} + a \left(- 1\right) + b = 3 \\ {0}^{2} + \frac{2}{3} {0}^{3} + a \cdot 0 + b = 2\end{matrix}\right.$

Solving for $a , b$ we have

$a = - \frac{2}{3} , b = 2$ so the curve is

$y = \frac{2}{3} {x}^{3} + {x}^{2} - \frac{2}{3} x + 2$

Mar 29, 2017

$y \left(x\right) = - \frac{2}{3} {x}^{3} + {x}^{2} + \frac{2}{3} x + 2$

#### Explanation:

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

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 - 4 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \int \left(2 - 4 x\right) \mathrm{dx} = 2 x - 2 {x}^{2} + {C}_{1}$

$y = \int \left(2 x - 2 {x}^{2} + {C}_{1}\right) \mathrm{dx} = {x}^{2} - \frac{2}{3} {x}^{3} + {C}_{1} x + {C}_{2}$

We can now find the values of ${C}_{1}$ and ${C}_{2}$ from the equations we get from the initial conditions:

$y \left(- 1\right) = 3$

$y \left(0\right) = 2$

so that:

$\left\{\begin{matrix}1 + \frac{2}{3} - {C}_{1} + {C}_{2} = 3 \\ {C}_{2} = 2\end{matrix}\right.$

$\left\{\begin{matrix}{C}_{1} = - 3 + 2 + 1 + \frac{2}{3} = \frac{2}{3} \\ {C}_{2} = 2\end{matrix}\right.$

and we can conclude that:

$y \left(x\right) = - \frac{2}{3} {x}^{3} + {x}^{2} + \frac{2}{3} x + 2$