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pav-90 [236]
3 years ago
8

The equation3y′=3xy+2x2+2y2x2, (∗)can be written in the form y′=f(y/x), i.e., it is homogeneous, so we can use the substitution

u=y/x to obtain a separable equation with dependent variable u=u(x). Introducing this substitution and using the fact that y′=xu′+u we can write (∗) asy′=xu′+u=f(u)where f(u)=Separating variables we can write the equation in the formg(u)du=dxxwhere g(u)= . An implicit general solution with dependent variable u can be written in the formln(x)−=CTransforming u=y/x back into the variables x and y and using the initial condition y(1)=1 we findC=Finally solve for y to obtain the explicit solution of the initial value problemy=

Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

The explicit solution of the initial value problem is y = xtan(\frac{2}{3}lnx +\frac{\pi}{4}  )

Step-by-step explanation:

The step by step explanation is shown on the first uploaded image

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Helpp please!!!
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Answer:

  h, j2, f, g, j1, i, k, l (ell)

Step-by-step explanation:

The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).

We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).

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  • j2(x): 3x^2/-x^2 = -3
  • f(x): 0x^2/(12x^2) = 0
  • k(x): 5x^2/x^2 = 5

So, the ordering least-to-greatest is ...

  h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)

6 0
3 years ago
The coordinates of the vertices of a triangle are A(-1, 2), B(3, 1), and C(-3, -2). The coordinates of the vertices of its image
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Answer:

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Step-by-step explanation:

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So, the transformation is (x, y) ⇒ (x +4, y -4).

Adding these values to a coordinate pair causes it to be translated to the right 4 units (x is increased by 4) and down 4 units (y is decreased by 4).

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