How to know if two triangles are congruent. Currently learning SAS, SSS, ASA, and AAS.
1 answer:
You might be interested in
Answer:
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 =