91Ó°ÊÓ

We need to estimate the value of the function \(f(t)\) at \(t=1\) using Euler's method, then find the explicit solution to the differential equation and compare the results.

Euler's method updates the solution using the formula: y_{n+1} = y_n + h \times f(t_n, y_n)where \(h\) is the step size, \(n\) is the number of steps, and \(f(t, y)\) is the derivative function.

Since we need to estimate \(f(1)\) with \(n=5\), the step size \(h\) can be calculated as: h = \frac{t_{\text{end}} - t_{\text{start}}}{n} = \frac{1-0}{5} = 0.2

Using \(h=0.2\) and \(y'=-(t+1)y^2\), we iteratively compute the values of \(y\).y_0 = y(0) = 1t_0 = 0t_1 = t_0 + h = 0.2y_1 = y_0 + h \times (- (t_0 + 1) \times y_0^2)= 1 + 0.2 \times (- (0 + 1) \times 1^2) = 1 + 0.2 \times (-1) = 0.8Continuing to iterate:t_2 = t_1 + h = 0.4y_2 = y_1 + h \times (-(t_1 + 1) \times y_1^2) = 0.8 + 0.2 \times (-(0.2 + 1) \times 0.8^2) = 0.6672t_3 = t_2 + h = 0.6y_3 = y_2 + h \times (-(t_2 + 1) \times y_2^2) = 0.6672 + 0.2 \times (-(0.4 + 1) \times 0.6672^2) = 0.5713t_4 = t_3 + h = 0.8y_4 = y_3 + h \times (-(t_3 + 1) \times y_3^2) = 0.5713 + 0.2 \times (-(0.6 + 1) \times 0.5713^2) = 0.4999t_5 = t_4 + h = 1y_5 = y_4 + h \times (-(t_4 + 1) \times y_4^2) = 0.4999 + 0.2 \times (-(0.8 + 1) \times 0.4999^2) = 0.4464Thus, the Euler's method estimate for \(f(1) = y_5\) is approximately 0.4464

To solve the differential equation analytically, integrate it:dy /( -y^2) = (t+1) dtIntegrate both sides:dy /( -y^2) = ∫ (-(y^{-2})) dy = ∫ (t + 1) dt simplifies to y^{-1} = (-\frac{(t-1)t}{2}) + cUsing initial condition \( y(0) = 1 \) 1 = c \therefore \frac(1){y} = (-\frac(t^2){2}+ 1-t). Rearrange to get: y(t)= (2c)/(2c-(t^{2}-2t)),Where c=1/2 substituting we get= y = \frac {1} {1+t^2/2-1}) the substituting value 1 we get y(1)=2

Using Euler's method, \(f(1) \rightarrow 0.4464\) ; Using explicit solution method \(f(1)\rightarrow 2\)