Respuesta :
Answer:
[tex]y\left(1.4\right)=0.992[/tex].
Step-by-step explanation:
The Euler's method states that [tex]y_{n+1}=y_n+h \cdot f \left(x_n, y_n \right)[/tex], where [tex]x_{n+1}=x_n + h[/tex].
To find [tex]y\left(1.4 \right)[/tex] for [tex]y'=- 4 x y + 4 x[/tex] when [tex]y\left(1 \right)=0[/tex], with step size [tex]h=0.2[/tex] using the Euler's method you must:
We have that [tex]h=0.2=\frac{1}{5}[/tex], [tex]x_0=1[/tex], [tex]y_0=0[/tex], [tex]f(x,y)=- 4 x y + 4 x[/tex].
Step 1.
[tex]x_{1}=x_{0}+h=1+\frac{1}{5}=\frac{6}{5}[/tex]
[tex]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[/tex]
Step 2.
[tex]x_{2}=x_{1}+h=\frac{6}{5}+\frac{1}{5}=\frac{7}{5}=1.4[/tex]
[tex]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[/tex]
The answer is [tex]y\left(1.4\right)=0.992[/tex]
