Question

5 Solve the variable separable x^3dx+(y+1)^2dy=0
(a) 3x^4+4(y+2)^3=C
(b) 5x^4+4(y+1)^3=C
© 3x^4+4(y+1)^3=C
(d) 4x^4+4(y+2)^3=C
6 Solve : (x^3+y^3)dx−3xy^2dy=0
(a) x^5−2y^3=Cx
(b) x^3−2y^3=Cx
© x^3−3y^3=Cx
(d) x^3−2y^2=Cx
7 Solve : (1+2e^x/y)dx+2e^x/y(1− x/y)dy=0
(a) 5x+2ye^x/y=C
(b) x+2ye^2x/y=C
© x+2ye^x/y=C
(d) 5x+3ye^x/y=C

Answer 1

ANSWER on QUESTION #59041 - Math - Differential Equations

QUESTION #1

Solve the variable separable $x^3 dx + (y + 1)^2 dy = 0$

a) $3x^4 + 4(y + 2)^3 = C$

b) $5x^4 + 4(y + 1)^3 = C$

c) $3x^4 + 4(y + 1)^3 = C$

d) $4x^4 + 4(y + 2)^3 = C$

SOLUTION

$$\begin{array}{l} x^3 dx + (y + 1)^2 dy = 0 \Leftrightarrow d\left(\frac{x^4}{4}\right) + d\left(\frac{(y + 1)^3}{3}\right) = 0 \Leftrightarrow \\ d\left(\frac{x^4}{4} + \frac{(y + 1)^3}{3}\right) = 0 \Leftrightarrow \frac{x^4}{4} + \frac{(y + 1)^3}{3} = Const \Leftrightarrow \\ 12 * \left(\frac{x^4}{4} + \frac{(y + 1)^3}{3}\right) = \underbrace{12 * Const}_{C} \Leftrightarrow 3x^4 + 4(y + 1)^3 = C \\ \end{array}$$

ANSWER: c) $3x^4 + 4(y + 1)^3 = C$

QUESTION #2

Solve $(x^3 + y^3)dx - 3xy^2 dy = 0$

a) $x^5 - 2y^3 = Cx$

b) $x^3 - 2y^3 = Cx$

c) $x^3 - 3y^3 = Cx$

d) $x^5 - 2y^2 = Cx$

SOLUTION

$$(x^3 + y^3)dx - 3xy^2 dy = 0$$ $$(x^3 + y^3) - 3xy^2 \frac{dy}{dx} = 0$$

Divide both sides by $x^3$:

$$1 + \frac {y ^ {3}}{x ^ {3}} - 3 \frac {x y ^ {2}}{x ^ {3}} \frac {d y}{d x} = 0$$

Let

$$\frac {y ^ {3}}{x ^ {3}} = z (x) \Leftrightarrow \frac {3 y ^ {2}}{x ^ {3}} \frac {d y}{d x} - \frac {3 y ^ {3}}{x ^ {4}} = \frac {d z}{d x} \Leftrightarrow \frac {3 y ^ {2}}{x ^ {3}} \frac {d y}{d x} = \frac {3 y ^ {3}}{x ^ {4}} + \frac {d z}{d x} = \frac {3 z}{x} + \frac {d z}{d x}$$ $$1 + \frac {y ^ {3}}{x ^ {3}} - x \frac {3 y ^ {2}}{x ^ {3}} \frac {d y}{d x} = 0 \Leftrightarrow 1 + z - x \left(\frac {3 z}{x} + \frac {d z}{d x}\right) = 0 \Leftrightarrow 1 + z - 3 z - x \frac {d z}{d x} = 0$$ $$1 - 2 z = x \frac {d z}{d x} \Leftrightarrow \frac {d x}{x} = \frac {d z}{1 - 2 z} \Leftrightarrow \ln | x | + \ln C = \frac {- 1}{2} \ln | 1 - 2 z | \Leftrightarrow$$ $$- 2 \ln | C x | = \ln \left| 1 - 2 \frac {y ^ {3}}{x ^ {3}} \right| \Leftrightarrow \ln | C x | ^ {- 2} = \ln \left| 1 - 2 \frac {y ^ {3}}{x ^ {3}} \right| \Leftrightarrow \ln \frac {1}{| C x | ^ {2}} = \ln \left| 1 - 2 \frac {y ^ {3}}{x ^ {3}} \right|$$ $$\frac {1}{| C x | ^ {2}} = 1 - 2 \frac {y ^ {3}}{x ^ {3}} \quad \text {multiplying by } x ^ {3} \Leftrightarrow \frac {x ^ {3}}{| C x | ^ {2}} = x ^ {3} - 2 y ^ {3} \Leftrightarrow \frac {1}{\frac {| C |}{C _ {1}}} x = x ^ {3} - 2 y ^ {3} \Leftrightarrow$$ $$x ^ {3} - 2 y ^ {3} = C _ {1} x$$

**ANSWER:** b) $x^{3} - 2y^{3} = Cx$

QUESTION #3

Solve $\left(1 + 2e^{\frac{x}{y}}\right)dx + 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0$

a) $5x + 2ye^{\frac{x}{y}} = C$

b) $x + 2ye^{\frac{2x}{y}} = C$

c) $x + 2ye^{\frac{x}{y}} = C$

d) $5x + 3ye^{\frac{x}{y}} = C$

SOLUTION

We write this differential equation in the form $P(x,y)dx + Q(x,y)dy = 0$, where

$P(x,y) = \left(1 + 2e^{\frac{x}{y}}\right)$ and $Q(x,y) = 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)$. We show that this equation is an exact ordinary differential equation, to do this, check the equality

$$\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$$ $$\frac {\partial P}{\partial y} = \frac {\partial}{\partial y} \left(1 + 2 e ^ {\frac {x}{y}}\right) = 2 e ^ {\frac {x}{y}} * \frac {- x}{y ^ {2}}$$ $$\frac {\partial Q}{\partial x} = \frac {\partial}{\partial x} \left(2 e ^ {\frac {x}{y}} \left(1 - \frac {x}{y}\right)\right) = 2 e ^ {\frac {x}{y}} * \frac {1}{y} \left(1 - \frac {x}{y}\right) + 2 e ^ {\frac {x}{y}} \left(- \frac {1}{y}\right) = 2 e ^ {\frac {x}{y}} * \frac {1}{y} \left(1 - \frac {x}{y} - 1\right) = 2 e ^ {\frac {x}{y}} * \frac {- x}{y ^ {2}}$$

As we can see

$$\frac {\partial P}{\partial y} = 2 e ^ {\frac {x}{y}} * \frac {- x}{y ^ {2}} = \frac {\partial Q}{\partial x}$$

This equality confirms our guess that this ordinary differential equation is exact. Therefore,

$$\begin{array}{l} P (x, y) = F _ {x} ^ {\prime} (x, y) \quad \text{and} \quad Q (x, y) = F _ {y} ^ {\prime} (x, y) \\ F _ {x} ^ {\prime} (x, y) = 1 + 2 e ^ {\frac {x}{y}} \Leftrightarrow F (x, y) = \int \left(1 + 2 e ^ {\frac {x}{y}}\right) d x = x + 2 y e ^ {\frac {x}{y}} + \varphi (y) \\ Q (x, y) = F _ {y} ^ {\prime} (x, y) = \frac {\partial}{\partial y} \left(x + 2 y e ^ {\frac {x}{y}} + \varphi (y)\right) = \varphi^ {\prime} (y) + 2 e ^ {\frac {x}{y}} + 2 y e ^ {\frac {x}{y}} * \frac {- x}{y ^ {2}} = \\ = \varphi^ {\prime} (y) + 2 e ^ {\frac {x}{y}} \left(1 - \frac {x}{y}\right) \Leftrightarrow \underbrace {2 e ^ {\frac {x}{y}} \left(1 - \frac {x}{y}\right)} _ {Q (x, y)} = \varphi^ {\prime} (y) + 2 e ^ {\frac {x}{y}} \left(1 - \frac {x}{y}\right) \Leftrightarrow \varphi^ {\prime} (y) = 0 \\ \varphi (y) = C _ {1} \\ P (x, y) d x + Q (x, y) d y = 0 \Leftrightarrow d F (x, y) = 0 \Leftrightarrow F (x, y) = C \\ x + 2 y e ^ {\frac {x}{y}} + C _ {1} = C \Leftrightarrow x + 2 y e ^ {\frac {x}{y}} = \underbrace {(C - C _ {1})} _ {C _ {2}} \Leftrightarrow x + 2 y e ^ {\frac {x}{y}} = C _ {2} \\ \end{array}$$

**ANSWER**: c) $x + 2ye^{\frac{x}{y}} = C$.

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