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 =

h

step-by-step explanation:

answer: there are x+184 pages in olivia's book.

step-by-step explanation:

let the total number of pages in the book which olive has already read be x.

then if olive has 184 pages left to read, then the total number of pages are in olive's book= pages she has read+pages she has left to read

therefore, the total number of pages in olivia's book=

hence, there are x+184 pages in olivia's book.