Apply Euler's method with the step h = 1/2 to find an approximation of y(1) where y is the solution of the initial value problem y'=1+y y(0)=0

Question

Apply Euler's method with the step h = 1/2 to find an approximation of y(1) where y is the solution of the initial value problem y'=1+y  y(0)=0

anonymous · Answer

i get the answer y approximately equals 7/2...can anyone verify?

anonymous · Answer

$$y(1)=e^1-1=1.71828 $$

anonymous · Answer

I haven't done Euler's method in a long time. But robtobey's answer doesn't look right to me! I will write down what I have.

anonymous · Answer

Let $f(y,t)=1+y$, and we have $t_0=0$ and $y_0=0$. Since $h=\frac{1}{2}$, we need to do two iterations in order to find $y(1)$.
First iteration (finding $y_1$ which is an approximation of y(0.5):
$f_0=f(0,0)=1\implies y_1=y_0+hf_0=\frac{1}{2}$.
Second iteration (Finding $y_2$, an approximation of $y(1)$):
$f_1=f(\frac{1}{2},\frac{1}{2})=1+\frac{1}{2}=\frac{3}{2} \implies y_2=y_1+hf_1=\frac{1}{2}+\frac{1}{2}\times\frac{3}{2}=\frac{5}{4}$.
So, $y(1)\approx \frac{5}{4}$.

anonymous · Answer

Do you have the final answer?

anonymous · Answer

The actual solution to the equation is attached and is evaluated at y(1).

anonymous · Answer

Yeah, it's very easy to find the exact solution from the auxiliary equation $m-1=0 \implies m=1$, which gives the solution associated with the  homogeneous equation to be $y_c=ce^x$. Its particular solution is some constant, call it $y_p=A$. Plugging it in the equation yields $0=1+A \implies A=-1$. So, the general solution is $y=y_c+y_p=ce^x-1$. And by using the given initial value, we have $0=c-1 \implies c=1$. So, $y=e^x-1$. Therefore, $y(1)=e-1$.