Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Round your results to four decimal places.
-$y ^ { \prime } = y ^ { 2 } ( 1 - 2 x ) , y ( - 1 ) = - 1 , d x = 0.5$
A) $\mathrm { y } _ { 1 } = 0.4 , \mathrm { y } _ { 2 } = 0.63 , \mathrm { y } _ { 3 } = 0.7472$
B) $\mathrm { y } _ { 1 } = 0.7 , \mathrm { y } _ { 2 } = 0.99 , \mathrm { y } _ { 3 } = 1.9136$
C) $\mathrm { y } _ { 1 } = 0.5 , \mathrm { y } _ { 2 } = 0.75 , \mathrm { y } _ { 3 } = 1.03125$
D) $\mathrm { y } _ { 1 } = 0.7 , \mathrm { y } _ { 2 } = 0.99 , \mathrm { y } _ { 3 } = 1.2656$

Question

Quizplus · Accepted Answer

Using Euler's method, we have:

\begin{aligned} y_{1} &= y_{0} + f(x_{0},y_{0})\cdot \Delta x \ &= -1 + (-1)^2(1-2(-1))\cdot 0.5 \ &= 0.5 \ y_{2} &= y_{1}+f(x_{1},y_{1})\cdot \Delta x \ &= 0.5 + 0.5^2(1-2(-0.5))\cdot 0.5 \ &= 0.75 \ y_{3} &= y_{2}+f(x_{2},y_{2})\cdot \Delta x \ &= 0.75 + 0.75^2(1-2(0))\cdot 0.5 \ &= 1.03125 \end{aligned}

Rounding to four decimal places, we get the first three approximations to be:

\begin{aligned} y_{1} &= 0.5 \ y_{2} &= 0.75 \ y_{3} &= 1.03125 \end{aligned}

Therefore, the best choice is C.

Use Euler's Method to Calculate the First Three Approximations to the Given