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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Covert 3.92 into a mixed number in simplest form?

12345 [234]

3.92

= 1.00 + 1.00 + 1.00 + 0.92

= 100/100 + 100/100 + 100/100 + 92/100

= 300/100 + 92/100

= 392/100

-------------------------------------------------------

392/100 = 196/50 = 98/25

3 0

3 years ago

Read 2 more answers

PLEASE HELP ILL GIVE MEDALS AND MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!! NEEDS TO BE ALEGABRA 2 MEATHOD!!!!!!!!

mars1129 [50]

Hi there,
This is the original inequality equation:
$\frac{x}{x+1} \ \textless \ \frac{x}{x-1}$
So, we first need to find the critical points of equality, and we can do that by switching the less than sign to an equal sign.
$\frac{x}{x+1} = \frac{x}{x-1}$
Now, we multiply both sides by x + 1:
$x= \frac{x^{2} +x}{x-1}$
Then, we multiply both sides by x - 1:
$x^{2} -x= x^{2} +x$
Next, we subtract x² from both sides:
$-x=x$
After that, we solve for x. We do this by adding -x to both sides and dividing by 2. Doing so gives us x = 0, which is our first critical point. We need to find a few more critical points by testing x = -1 and x = 1. Here is how we do that:
x = −1 (Makes left denominator equal to 0)x = 1 (Makes right denominator equal to 0)Check intervals in between critical points. (Test values in the intervals to see if they work.)x <−1 (Doesn't work in original inequality)−1 < x <0 (Works in original inequality)0 < x < 1 (Doesn't work in original inequality)x > 1 (Works in original inequality)
Therefore, the answer to your query is -1 < x < 0 or x > 1. Hope this helps and have a phenomenal day!

4 0

3 years ago

Laura is building a model for her class. She is using a scale of 2 centimeters represents 5 feet. If the building is 55 feet lon

Leona [35]

22 i think the right answer

4 0

3 years ago

I need help asap i been stuck on this for a long time

Talja [164]

Answer:

1 to 2 is an example inches to 2 stuck etc

5 0

3 years ago

Read 2 more answers