Find out the answer to "Solve the following IVP for the exact equation..."

widdonod1t

Answered question

2021-12-29

Solve the following IVP for the exact equation and find the interval of validity for the solution.

2 x y - 9 x^{2} + (2 y + x^{2} + 1) \frac{d y}{d x} = 0 y (0) = - 2

Answer & Explanation

Barbara Meeker

Beginner2021-12-30Added 38 answers

For any differential equation that is given to be exact, we must have a function f(x,y) such that:

d [f (x, y)] = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = M d x + N d y

∴ M (x, y) = \frac{\partial f}{\partial x} and N (x, y) = \frac{\partial f}{\partial y}

i . e . f (x, y) = \int M (x, y) d x + ϕ_{1} (y) and f (x, y) = \int N (x, y) d y + ϕ_{2} (x)

The given differential equation can be written as:

2 x y - 9 x^{2} + (2 y + x^{2} + 1) \frac{d y}{d x} = 0

(2 x y - 9 x^{2}) d x + (2 y + x^{2} + 1) d y = 0

(i)
Comparing to the standard form of

M (x, y) d x + N (x . y) d y = 0

we get:

M (x, y) = 2 x y - 9 x^{2}

∴ f (x, y) = \int (2 x y - 9 x^{2}) d x + ϕ_{1} (y)

= 2 y (\frac{x^{2}}{2}) - 9 (\frac{x^{3}}{3}) + ϕ_{1} (y)

= x^{2} y - 3 x^{3} + ϕ_{1} (y)

...(2)

N (x, y) = 2 y + x^{2} + 1

∴ f (x, y) = \int (2 y + x^{2} + 1) d y + ϕ_{2} (x)

= 2 (\frac{y^{2}}{2}) + x^{2} (y) + (y) + ϕ_{2} (x)

reinosodairyshm Beginner2021-12-31Added 36 answers

[y (2 x) d x + x^{2} (d y)] + (2 y + 1) d y - 9 x^{2} d x = 0

d (x^{2} y) + d (y^{2} + y) - d (3 x^{3}) = 0

\Rightarrow d (x^{2} y + y^{2} + y - 3 x^{2}) = 0

\Rightarrow x^{2} y + y^{2} + y - 3 x^{2} = c

y (0) = - 3 \Rightarrow 0 + 4 - 3 + 0 = c = 6 \Rightarrow x^{2} y + y^{2} + y - 3 x^{2} = 6

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} (2 x y - 9 x^{2}) d x + (2 y + x^{2} + 1) d y = 0 \\ M_{y} = 2 x = N_{x} = 2 x \\ \int M d x + g (y) \\ H (x, y) = \int 2 x y - 9 x^{2} d x + g (y) = \int 2 y + x^{2} + 1 d y \\ = x^{2} y - 3 x^{3} + g (y) = y^{2} + x^{2} y + y \end{matrix}$
Since $x^{2} y$ appears on both sides, cancel it on the yside, which gives $g (y) = y^{2} + y$ . Add this to the leftside and the general solution is:
$H (x, y) = x^{2} y - 3 x^{2} + y^{2} + y$

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