Respuesta :
Answer:
- [tex]y(0.2)=3[/tex], [tex]y(0.4)=3.005974448[/tex], [tex]y(0.6)=3.017852169[/tex], [tex]y(0.8)=3.035458382[/tex], and [tex]y(1.0)=3.058523645[/tex]
- The general solution is [tex]y=\ln \left(\frac{3t^2}{2}+e^3\right)[/tex]
- The error in the approximations to y(0.2), y(0.6), and y(1):
[tex]|y(0.2)-y_{1}|=0.002982771[/tex]
[tex]|y(0.6)-y_{3}|=0.008677796[/tex]
[tex]|y(1)-y_{5}|=0.013499859[/tex]
Step-by-step explanation:
Point a:
The Euler's method states that:
[tex]y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)[/tex] where [tex]t_{n+1}=t_n + h[/tex]
We have that [tex]h=0.2[/tex], [tex]t_{0}=0[/tex], [tex]y_{0} =3[/tex], [tex]f(t,y)=3te^{-y}[/tex]
- We need to find [tex]y(0.2)[/tex] for [tex]y'=3te^{-y}[/tex], when [tex]y(0)=3[/tex], [tex]h=0.2[/tex] using the Euler's method.
So you need to:
[tex]t_{1}=t_{0}+h=0+0.2=0.2[/tex]
[tex]y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=[/tex]
[tex]=3 + 0.2 \cdot \left(0 \right)= 3[/tex]
[tex]y(0.2)=3[/tex]
- We need to find [tex]y(0.4)[/tex] for [tex]y'=3te^{-y}[/tex], when [tex]y(0)=3[/tex], [tex]h=0.2[/tex] using the Euler's method.
So you need to:
[tex]t_{2}=t_{1}+h=0.2+0.2=0.4[/tex]
[tex]y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=[/tex]
[tex]=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448[/tex]
[tex]y(0.4)=3.005974448[/tex]
The Euler's Method is detailed in the following table.
Point b:
To find the general solution of [tex]y'=3te^{-y}[/tex] you need to:
Rewrite in the form of a first order separable ODE:
[tex]e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt[/tex]
Integrate each side:
[tex]\int \:e^ydy=e^y+C[/tex]
[tex]\int \:3t\:dt=\frac{3t^2}{2}+C[/tex]
[tex]e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}[/tex]
We know the initial condition y(0) = 3, we are going to use it to find the value of [tex]C_{1}[/tex]
[tex]e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3[/tex]
So we have:
[tex]e^y=\frac{3t^2}{2}+e^3[/tex]
Solving for y we get:
[tex]\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)[/tex]
Point c:
To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:
Find the values y(0.2), y(0.6), and y(1) using [tex]y=\ln \left(\frac{3t^2}{2}+e^3\right)[/tex]
[tex]y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771[/tex]
[tex]y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965[/tex]
[tex]y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504[/tex]
Next, where [tex]y_{1}, y_{3}, \:and \:y_{5}[/tex] are from the table.
[tex]|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771[/tex]
[tex]|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796[/tex]
[tex]|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859[/tex]
