Suppose $\frac { d y } { d x } = y ^ { 2 } - 3$ , $y ( 1 ) = 2$ . Use Euler's method with step size $h = 1$ to approximate $y ( 3 )$ .
A)1
B)8
C)16
D)9
E)4
F)14
G)12
H)6

Question

Quizplus · Accepted Answer

Euler's method with step size $h = 1$ starts with $y(1) = 2$. For the first step to approximate $y(2)$, use $\frac{dy}{dx} = y^2 - 3$ with $y(1) = 2$, so $\frac{dy}{dx} = 2^2 - 3 = 1$. Thus, $y(2) = y(1) + h \cdot \frac{dy}{dx} = 2 + 1 \cdot 1 = 3$. For the next step to approximate $y(3)$, use $y(2) = 3$, so $\frac{dy}{dx} = 3^2 - 3 = 6$. Thus, $y(3) = y(2) + h \cdot \frac{dy}{dx} = 3 + 1 \cdot 6 = 9$.

Suppose dydx=y2−3\frac { d y } { d x } = y ^ { 2 } - 3dxdy​=y2−3

Suppose $\frac { d y } { d x } = y ^ { 2 } - 3$