Use Euler's method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05. Find an explicit solution for the initial-value problem. y â²=y, y(0)=1; y(1.0)

y(1) = 2.5937424601 by using h = 0.1

y(1)=2.65329770514442 by using h = 0.05

Explicit solution - y =

Step-by-step explanation:

Euler's Method -  yₙ₊₁=yₙ+h⋅f(tₙ,yₙ), where tₙ₊₁=tₙ+h.

We have that h=0.1 , t₀=0, y₀=1, f(t,y)=y

Step 1 -

t₁=t₀+h=0+0.1=0.1

y(t₁)=y(0.1)=y₁=y₀+h⋅f(t₀,y₀)

                    =1+h⋅f(0,1)

                    =1+(0.1)⋅(1.0)

                     =1.1

Step 2.-

t₂=t₁+h=0.1+0.1=0.2

y(t₂)=y(0.2)=y₂=y₁+h⋅f(t₁,y₁)

                      =1.1+h⋅f(0.1,1.1)

                      =1.1+(0.1)⋅(1.1)

                      =1.21

Step 3-

t₃=t₂+h=0.2+0.1=0.3

y(t₃)=y(0.3)=y₃=y₂+h⋅f(t₂,y₂)

                      =1.21+h⋅f(0.2,1.21)

                      =1.21+(0.1)⋅(1.21)

                      =1.331

Step 4 -

t₄=t₃+h=0.3+0.1=0.4

y(t₄)=y(0.4)=y₄=y₃+h⋅f(t₃,y₃)

                      =1.331+h⋅f(0.3,1.331)

                      =1.331+(0.1)⋅(1.331)

                      =1.4641

Step 5.-

t₅=t₄+h=0.4+0.1=0.5

y(t₅)=y(0.5)=y₅=y₄+h⋅f(t₄,y₄)

                      =1.4641+h⋅f(0.5,1.4641)

                      =1.4641+(0.1)⋅(1.4641)

                      =1.61051

Step 6-

t₆=t₅+h=0.5+0.1=0.6

y(t₆)=y(0.6)=y₆=y₅+h⋅f(t₅,y₅)

                      =1.61051+h⋅f(0.5,1.61051)

                      =1.61051+(0.1)⋅(1.61051)

                      =1.771561

Step 7-

t₇=t₆+h=0.6+0.1=0.7

y(t₇)=y(0.7)=y₇=y₆+h⋅f(t₆,y₆)

                      =1.771561+h⋅f(0.6,1.771561)

                      =1.771561+(0.1)⋅(1.771561)

                      =1.9487171

Step 8-

t₈=t₇+h=0.7+0.1=0.8

y(t₈)=y(0.8)=y₈=y₇+h⋅f(t₇,y₇)

                      =1.9487171+h⋅f(0.7,1.9487171)

                      =1.9487171+(0.1)⋅(1.9487171)

                      =2.14358881

Step 9-

t₉=t₈+h=0.8+0.1=0.9

y(t₉)=y(0.9)=y₉=y₈+h⋅f(t₈,y₈)

                      =2.14358881+h⋅f(0.8,2.14358881)

                      =2.14358881+(0.1)⋅(2.14358881)

                      =2.357947691

Step 10.-

t₁₀=t₉+h=0.9+0.1=1.0

y(t₁₀)=y(1.0)=y₁₀=y₉+h⋅f(t₉,y₉)

                =2.357947691+h⋅f(0.9,2.357947691)

                =2.357947691+(0.1)⋅(2.357947691)

                =2.5937424601

∴ we get

y(1) = 2.5937424601 by using h = 0.1

same process goes with h = 0.05

we get y(1)=2.65329770514442

Now,

For the explicit solution  -

y' = y

⇒

⇒

By integrate, we get

∫ = ∫dx

⇒ln(y) = x

⇒y =

45.36 feet per second.

step-by-step explanation:

to find the most accurate determination for the number of feet per second tyrone traveled on his way to work let us convert miles to feet and minutes to seconds.

1 mile = 5280 feet.

1 minute = 60 seconds.

now we will divide 158400 by 3492 to find the most accurate determination for the number of feet per second tyrone traveled on his way to work.

therefore, tyrone traveled on his way to work at the rate of 45.36 feet per second.

35/2 is it i think maybe