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Likurg_2 [28]
3 years ago
15

We are given a differential equation and will rewrite it in the form M(x, y) dx + N(x, y) dy = 0. xy2 dy dx = y3 − x3 xy2 dy = (

y3 − x3) dx xy2 dy − (y3 − x3) dx = 0 xy2 dy + (−y3 + x3) dx = 0
Mathematics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

Check attachment for orderliness of question

Step-by-step explanation:

We want to the equation in an exact form

M(x, y) dx + N(x, y) dy = 0.

xy² dy/dx = y³− x³

Cross multiply

Then,

xy²dy=(y³-x³)dx

xy²dy - (y³-x³)dx = 0

xy²dy + (x³-y³)dx=0

Therefore,

(x³-y³)dx + xy²dy=0

Can only be an exact equation if and only if

dM/dy=dN/dx

From the equation

M=(x³-y³)

Then, dM/dy=-3y²

Also, N=xy²

dN/dx=y²

Since dM/dy ≠ dN/dx

Then it is not an examct equation and we can say

M≠(x³-y³)

And

N≠(xy²)

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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.

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