Question

Solve dy/dx = y^3+3x^2y ÷ x^3+3xy^2

Answer 1

Answer on Question #60659 - Math - Differential Equations

Question

Solve

$$\mathrm {d} y / \mathrm {d} x = - y ^ {\wedge} 3 + 3 x ^ {\wedge} 2 y \div x ^ {\wedge} 3 + 3 x y ^ {\wedge} 2$$

Solution

Method 1

Given

$$\frac {d y}{d x} = - \frac {y ^ {3} + 3 x ^ {2} y}{x ^ {3} + 3 x y ^ {2}},$$

divide the numerator and the denominator in the right-hand side of equation (1) by $x^3$

Then

$$\frac {d y}{d x} = - \frac {\left(\frac {y}{x}\right) ^ {3} + 3 \frac {y}{x}}{1 + 3 \left(\frac {y}{x}\right) ^ {2}}.$$

Substitute

$$u = u (x) = \frac {y}{x},$$

hence $y = u x,\frac{d y}{d x} = x\frac{d u}{d x} +u.$

In terms of $x$ and $u$ equation (2) takes the following form:

$$x \frac {d u}{d x} + u = - \frac {u ^ {3} + 3 u}{1 + 3 u},$$ $$x \frac {d u}{d x} = - \frac {u ^ {3} + 3 u}{1 + 3 u} - u,$$ $$x \frac {d u}{d x} = \frac {- u ^ {3} - 3 u - u - 3 u ^ {2}}{3 u + 1},$$ $$x \frac {d u}{d x} = - u \frac {u ^ {2} + 3 u + 4}{3 u + 1}$$ $$\frac {3 u + 1}{u \left(u ^ {2} + 3 u + 4\right)} = - \frac {d x}{x}$$

Equation $u^2 + 3u + 4 = 0$ does not have real solutions.

$$\frac {3 u + 1}{u (u ^ {2} + 3 u + 4)} = \frac {A}{u} + \frac {B u + C}{u ^ {2} + 3 u + 4}$$

Reduce the right-hand side to the common denominator

$$\frac {3 u + 1}{u (u ^ {2} + 3 u + 4)} = \frac {A (u ^ {2} + 3 u + 4) + u (B u + C)}{u (u ^ {2} + 3 u + 4)} = \frac {(A + B) u ^ {2} + (3 A + C) u + 4 A}{u (u ^ {2} + 3 u + 4)}$$

Numerators in the left-hand and in the right-hand sides are equal:

$$3 u + 1 = (A + B) u ^ {2} + (3 A + C) u + 4 A.$$

Equate the like terms and get the following system of equations:

$$\left\{ \begin{array}{c} A + B = 0, \\ 3 A + C = 3, \\ 4 A = 1, \end{array} \right.$$ $$\left\{ \begin{array}{c} B = - A, \\ C = 3 - 3 A, \\ A = \frac {1}{4}. \end{array} \right.$$ $$\left\{ \begin{array}{c} B = - \frac {1}{4}, \\ C = 3 - \frac {3}{4} = \frac {9}{4}, \\ A = \frac {1}{4}. \end{array} \right.$$

Thus,

$$\left(u ^ {2} + 3 u + 4\right) ^ {\prime} = 2 u + 3$$ $$\begin{array}{l} \frac {3 u + 1}{u \left(u ^ {2} + 3 u + 4\right)} = \frac {1}{4 u} + \frac {- u + 9}{4 \left(u ^ {2} + 3 u + 4\right)} = \frac {1}{4 u} + \frac {- u - \frac {3}{2} + 9 + \frac {3}{2}}{4 \left(u ^ {2} + 2 \cdot \frac {3}{2} u + \frac {9}{4} + 4 - \frac {9}{4}\right)} \\ = \frac {1}{4 u} - \frac {u + \frac {3}{2}}{4 \left(\left(u + \frac {3}{2}\right) ^ {2} + \frac {7}{4}\right)} + \frac {2 1}{2 \cdot 4 \left(\left(u + \frac {3}{2}\right) ^ {2} + \frac {7}{4}\right)} \\ = \frac {1}{4 u} - \frac {1}{2 \cdot 4} \cdot \frac {\left(u ^ {2} + 3 u + 4\right) ^ {\prime}}{\left(u ^ {2} + 3 u + 4\right)} + \frac {2 1}{8} \cdot \frac {1}{\left(u + \frac {3}{2}\right) ^ {2} + \frac {7}{4}} \\ \end{array}$$

Integrating both sides of (4)

$$\int \left(\frac {1}{4 u} - \frac {1}{8} \cdot \frac {(u ^ {2} + 3 u + 4) ^ {\prime}}{u ^ {2} + 3 u + 4} + \frac {2 1}{8} \cdot \frac {1}{\left(u + \frac {3}{2}\right) ^ {2} + \frac {7}{4}}\right) d u = - \int \frac {d x}{x}$$ $$\frac {1}{4} \ln | u | - \frac {1}{8} \ln (u ^ {2} + 3 u + 4) + \frac {2 1}{8} \cdot \frac {\frac {1}{\sqrt {7}}}{2} \tan^ {- 1} \frac {u + \frac {3}{2}}{\frac {\sqrt {7}}{2}} + C _ {1} = - \ln | x |,$$

where $\tan^{-1}$ is the inverse of the tangent, $C_1$ is an integration constant,

$$\frac {1}{4} \ln | u | - \frac {1}{8} \ln (u ^ {2} + 3 u + 4) + \frac {3 \sqrt {7}}{4} \tan^ {- 1} \frac {2 u + 3}{\sqrt {7}} + \ln | x | + C _ {1} = 0,$$ $$\frac {1}{4} \ln \frac {\left| u \right| \cdot \left| x \right| ^ {4}}{\sqrt {u ^ {2} + 3 u + 4}} + \frac {3 \sqrt {7}}{4} \tan^ {- 1} \frac {2 u + 3}{\sqrt {7}} + C _ {1} = 0,$$

multiplying (5) by 4 and substituting (3) into (5) obtain

$$\ln \frac {\left| \underline {{y}} \right| | x | ^ {4}}{\sqrt {\left(\underline {{x}}\right) ^ {2} + 3 \underline {{x}} + 4}} + 3 \sqrt {7} \cdot \tan^ {- 1} \frac {2 \underline {{y}} + 3}{\sqrt {7}} + C = 0,$$ $$\ln \frac {| y | | x | ^ {4}}{\sqrt {y ^ {2} + 3 y x + 4 x ^ {2}}} + 3 \sqrt {7} \cdot \tan^ {- 1} \frac {2 y + 3 x}{x \sqrt {7}} + C = 0$$

Method 2

$$\frac {d y}{d x} = - \frac {y ^ {3} + 3 x ^ {2} y}{x ^ {3} + 3 x y ^ {2}},$$

Rewrite the differential equation (6) in the following form:

$$\frac {d y}{d x} = - \frac {F _ {x}}{F _ {y}},$$

where $F_{x} = y^{3} + 3x^{2}y,F_{y} = x^{3} + 3xy^{2}$

Equation (7) can be rewritten as

$$F _ {x} d x + F _ {y} d y = 0.$$

If the function $F = F(x,y)$ is such a function that

$$\frac {\partial F}{\partial x} = F _ {x}, \frac {\partial F}{\partial y} = F _ {y},$$

then equation (8) becomes

$$\frac {\partial F}{\partial x} d x + \frac {\partial F}{\partial y} d y = 0,$$

that is,

$$d F (x, y) = 0,$$

hence $F(x,y) = C$ , where $C$ is an arbitrary real constant.

The rule for implicit differentiation in two variables is

$$\frac {d y}{d x} = - \frac {F _ {x}}{F _ {y}}$$ $$F _ {x} = (y ^ {3} + 3 x ^ {2} y) \rightarrow F = \int (y ^ {3} + 3 x ^ {2} y) d x = (x y ^ {3} + x ^ {3} y) + g (y),$$ $$F _ {y} = x ^ {3} + 3 x y ^ {2} = x ^ {3} + 3 x y ^ {2} + g ^ {\prime} (y),$$ $$g ^ {\prime} (y) = 0 \rightarrow g (y) = C,$$

where $C$ is an integration constant,

$$F (x, y) = x y ^ {3} + x ^ {3} y + C = 0.$$

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