Expert Answer to "Determine which of the following differential equations are..."

Cheexorgeny

Answered question

2021-12-31

Determine which of the following differential equations are homogeneous, then solve the general solution. If initial condition is given, find the particular solution:

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

Answer & Explanation

Jim Hunt

Beginner2022-01-01Added 45 answers

$x y d x - {(x + 2 y)}^{2} d y = 0$
$x y d x = {(x + 2 y)}^{2} d y$
$\frac{d x}{d y} = \frac{x^{2} + 4 y^{2} + 4 x y}{x y}$
$\frac{d x}{d y} = (\frac{x}{y}) + 4 (\frac{y}{x}) + 4$
It is a Homogeneous differential equation.
Let $\frac{x}{y} = v \Rightarrow x = v y$
So, $\frac{d x}{d y} = v + y \frac{d v}{d y}$
$∴ v + y \frac{d v}{d y} = v + \frac{4}{v} + 4$
$y \frac{d v}{d y} = 4 (\frac{v + 1}{v})$
$\frac{v d v}{v + 1} = 4 \frac{d y}{y}$
It becomes variable separable equation. So, now integrate both sides.
$\int \frac{v d v}{v + 1} = 4 \int \frac{d y}{y}$
$\int (\frac{v + 1 - 1}{v + 1}) d v = 4 \int \frac{d y}{y}$
$\int (1 - \frac{1}{v + 1}) d v = 4 \ln | y | + \ln c$
$v - \ln | v + 1 | = \ln (y^{4} c)$
$∴ \frac{x}{y} + \ln | \frac{x}{y} + 1 | = \ln | y^{4} c |$
$\frac{x}{y} = \ln | y^{4} c | - \ln | \frac{x + y}{y} |$
$\frac{x}{y} = \ln | \frac{(y^{4} c)}{(x +}$

Serita Dewitt Beginner2022-01-02Added 41 answers

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

x y d x = {(x + 2 y)}^{2} d y

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

\frac{d x}{d y} = \frac{x^{2} + 4 y^{2} + 4 x y}{x y}

\frac{d x}{d y} = \frac{x}{y} + 4 \frac{y}{x} + 4

.
Substituting;

x = y v

Diff. w.r.t. y

\frac{d x}{d y} = v + y \frac{d v}{d y}

v + y \frac{d v}{d y} = v + 4 \frac{1}{v} + 4

\neg {v} + y \frac{d v}{d y} = \neg {v} + 4 (\frac{1}{v} + 1)

\frac{y}{4} \frac{d v}{d y} = \frac{1 + v}{v}

Integrating

\int \frac{v}{v + 1} d v = 4 \int \frac{1}{y} d y

\int \frac{v + 1 - 1}{v + 1} d v = 4 \int \frac{1}{y} d y

v - \ln | v + 1 | = 4 \log y + c

applying

v = \frac{x}{y}

\frac{x}{y} - \ln | \frac{x}{y} + 1 | = 4 \log y + c

karton

Expert2022-01-10Added 613 answers

$\begin{matrix} x y d x - (x + 2 y)^{2} d y = 0 \\ x y d x = (x + 2 y)^{2} d y \\ \frac{d x}{d y} = \frac{x^{2} + 4 y^{2} + 4 x y}{x y} \\ \frac{d x}{d y} = (\frac{x}{y}) + 4 (\frac{y}{x}) + 4 \\ \frac{x}{y} = v \Rightarrow x = v y \\ \frac{d x}{d y} = v + y \frac{d v}{d y} \\ ∴ ⧸ v + y \frac{d v}{d y} = ⧸ v + \frac{4}{v} + 4 \\ y \frac{d v}{d y} = 4 (\frac{v + 1}{v}) \\ \frac{v d v}{v + 1} = 4 \frac{d y}{y} \\ \int \frac{v d v}{v + 1} = 4 \int \frac{d y}{y} \\ \int (\frac{v + 1 - 1}{v + 1}) d v = 4 \int \frac{d y}{y} \\ \int (1 - \frac{1}{v + 1}) d v = 4 \ln | y | + \ln c \\ v - \ln | v + 1 | = \ln (y^{4} c) \\ ∴ \frac{x}{y} + \ln | \frac{x}{y} + 1 | = \ln | y^{4} c | \\ \frac{x}{y} = \ln | y^{4} c | - \ln | \frac{x + y}{y} | \\ \frac{x}{y} = \ln | \frac{(y^{4} c)}{(\frac{x + y}{y})} | \\ x = y \ln | \frac{c y^{5}}{x + y} | \end{matrix}$

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