Question

(a) Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y' = 4x − 4xy, y(1) = 0. (Round your answer to four decimal places.) y(1.4) =

Answer 1

Answer:

$y\left(1.4\right)=0.992$.

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 $x_{n+1}=x_n + h$.

To find $y\left(1.4 \right)$ for $y'=- 4 x y + 4 x$ when $y\left(1 \right)=0$, with step size $h=0.2$ using the Euler's method you must:

We have that $h=0.2=\frac{1}{5}$, $x_0=1$, $y_0=0$, $f(x,y)=- 4 x y + 4 x$.

Step 1.

$x_{1}=x_{0}+h=1+\frac{1}{5}=\frac{6}{5}$

$y\left(x_{1}\right)=y\left( \frac{6}{5} \right)=y_{1}=y_{0}+h \cdot f \left(x_{0}, y_{0} \right)=0+h \cdot f \left(1, 0 \right)=0 + \frac{1}{5} \cdot \left(4.0 \right)=0.8$

Step 2.

$x_{2}=x_{1}+h=\frac{6}{5}+\frac{1}{5}=\frac{7}{5}=1.4$

$y\left(x_{2}\right)=y\left( \frac{7}{5} \right)=y_{2}=y_{1}+h \cdot f \left(x_{1}, y_{1} \right)=0.8+h \cdot f \left(\frac{6}{5}, 0.8 \right)=0.8 + \frac{1}{5} \cdot \left(0.96 \right)=0.992$

The answer is $y\left(1.4\right)=0.992$