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

Answer:

$x\left(\frac{7}{4}\right)=2.875$.

Step-by-step 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 $x(1.75)=x\left(\frac{7}{4}\right)$ for $\frac{dx}{dt}=2x$, when $t_0=1$, $x_0=1$, and $h=0.25$ using the Euler's method.

Step 1.

$t_{1}=t_{0}+h=1+\frac{1}{4}=\frac{5}{4}$

$x\left(t_{1}\right)=x\left( \frac{5}{4} \right)=x_{1}=x_{0}+h \cdot f \left(t_{0}, x_{0} \right)=1+h \cdot f \left(1, 1 \right)$

$x\left(t_{1}\right)=1 + \frac{1}{4} \cdot \left(2.0 \right)=1.5$

Step 2.

$t_{2}=t_{1}+h=\frac{5}{4}+\frac{1}{4}=\frac{3}{2}$

$x\left(t_{2}\right)=x\left( \frac{3}{2} \right)=x_{2}=x_{1}+h \cdot f \left(t_{1}, x_{1} \right)=1.5+h \cdot f \left(\frac{5}{4}, 1.5 \right)$

$x\left(x_{2}\right)=1.5 + \frac{1}{4} \cdot \left(2.5 \right)=2.125$

Step 3.

$t_{3}=t_{2}+h=\frac{3}{2}+\frac{1}{4}=\frac{7}{4}$

$x\left(t_{3}\right)=x\left( \frac{7}{4} \right)=x_{3}=x_{2}+h \cdot f \left(t_{2}, x_{2} \right)=2.125+h \cdot f \left(\frac{3}{2}, 2.125 \right)$

$x\left(t_{3}\right)=2.125 + \frac{1}{4} \cdot \left(3.0 \right)=2.875$