Question

Use Euler’s approximation to determine x(1.75), the value of a function at t = 1.75, given increments of time of 0.25, x(1) = 1, and dx/dt = 2x

Answer 1

`x\left((7)/(4)\right)=2.875`.

Explanation:

The Euler's method states that
`y_(n+1)=y_n+h \cdot f \left(x_n, y_n \right)`, where
`x_(n+1)=x_n + h`.

We want to find
`x(1.75)=x\left((7)/(4)\right)` for
`(dx)/(dt)=2x`, when
`t_0=1`,
`x_0=1`, and
`h=0.25` using the Euler's method.

Step 1.

`t_(1)=t_(0)+h=1+(1)/(4)=(5)/(4)`

`x\left(t_(1)\right)=x\left( (5)/(4) \right)=x_(1)=x_(0)+h \cdot f \left(t_(0), x_(0) \right)=1+h \cdot f \left(1, 1 \right)`

`x\left(t_(1)\right)=1 + (1)/(4) \cdot \left(2.0 \right)=1.5`

Step 2.

`t_(2)=t_(1)+h=(5)/(4)+(1)/(4)=(3)/(2)`

`x\left(t_(2)\right)=x\left( (3)/(2) \right)=x_(2)=x_(1)+h \cdot f \left(t_(1), x_(1) \right)=1.5+h \cdot f \left((5)/(4), 1.5 \right)`

`x\left(x_(2)\right)=1.5 + (1)/(4) \cdot \left(2.5 \right)=2.125`

Step 3.

`t_(3)=t_(2)+h=(3)/(2)+(1)/(4)=(7)/(4)`

`x\left(t_(3)\right)=x\left( (7)/(4) \right)=x_(3)=x_(2)+h \cdot f \left(t_(2), x_(2) \right)=2.125+h \cdot f \left((3)/(2), 2.125 \right)`

`x\left(t_(3)\right)=2.125 + (1)/(4) \cdot \left(3.0 \right)=2.875`