Question

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)

Answer 1

Answer:

y(1) = 2.5937424601 by using h = 0.1

y(1)=2.65329770514442 by using h = 0.05

Explicit solution - y = $e^{x}$

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

⇒$\frac{dy}{dx} = y$

⇒$\frac{dy}{y} = dx$

By integrate, we get

∫$\frac{dy}{y}$ = ∫dx

⇒ln(y) = x

⇒y = $e^{x}$