Expert Answer to "Find the solution of the following Differential Equationsy′=(y−x)"

Name: Find the solution of the following Differential Equationsy′=(y−x)
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the solution of the following Differential Equationsy′=(y−x)

reproacht3

Answered question

2021-12-26

Find the solution of the following Differential Equations

y^{'} = (y - x)

Answer & Explanation

puhnut1m

Beginner2021-12-27Added 33 answers

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

Let

y - x = t

\frac{d y}{d x} - 1 = \frac{d t}{d x}

\Rightarrow \frac{d y}{d x} = \frac{d t}{d x} + 1

\frac{d t}{d x} + 1 = t^{2}

\frac{d t}{d x} = t^{2} - 1

\int \frac{d t}{t^{2} - 1} = \int d x

\frac{1}{2} \ln (\frac{t - 1}{t + 1}) = x + c

\ln (\frac{y - x - 1}{y - x + 1}) = 2 x + 2 c

Jimmy Macias

Beginner2021-12-28Added 30 answers

y = x + \frac{a^{2} + x^{2}}{a^{2} - x^{2}}

where a is a constant of integration.
Explanation:
Let us change variables to

u = y - x

Then

u^{'} = y^{'} - 1

and so the differential equation becomes

u^{'} = u^{2} - 1 \Rightarrow \frac{d u}{u^{2} - 1} = d x \Rightarrow \frac{1}{2} \log | \frac{u - 1}{u + 1} | = \log x - \log a

where we have written the constant of integration as

- \log a

. Thus

\frac{u - 1}{u + 1} = \frac{x^{2}}{a^{2}} \Rightarrow u = \frac{a^{2} + x^{2}}{a^{2} - x^{2}}

karton

Expert2022-01-09Added 613 answers

Explanation:
We have: $y^{'} = (y - x)^{2} [A]$
Perform the substitution:
u=y-x
Then differentiating wrt x, and applying the product rule:
$\frac{d u}{d x} = \frac{d y}{d x} - 1 \Rightarrow \frac{d y}{d x} = 1 + \frac{d u}{d x}$
Substituting into the DE:
$1 + \frac{d u}{d x} = u^{2}$
$∴ \frac{d u}{d x} = u^{2} - 1$
Reducing the DE to a separable form:
$\frac{1}{u^{2} - 1} \frac{d u}{d x} = 1$
So we can ''separate the variables'' to get:
$\int \frac{1}{u^{2} - 1} d u = \int d x$
We can perform partial fraction decomposition of the LHS integrand:
$\frac{1}{u^{2} - 1} = \frac{1}{(u + 1) (u - 1)}$
$= \frac{A}{u + 1} + \frac{B}{u - 1}$
And using the ''cover up'' methods we get:
$A = - \frac{1}{2}, B = \frac{1}{2}$
Allowing us to write the equation as:
$\frac{1}{2} \int \frac{1}{u - 1} - \frac{1}{u + 1} d u = \int d x$
And integrating, we get:
$\begin{matrix} \frac{1}{2} \ln (u - 1) - \ln (u + 1) = x + C \\ \ln (\frac{u - 1}{u + 1}) = 2 x + 2 C \\ \frac{u - 1}{u + 1} = e^{2 x + 2 C} \\ u - 1 = (u + 1) A e^{2 x} \\ u - 1 = u A e^{2 x} + A e^{2 x} \\ u - u A e^{2 x} = 1 + A e^{2 x} \\ u (1 - A e^{2 x}) = 1 + A e^{2 x} \\ u = \frac{1 + A e^{2 x}}{1 - A e^{2 x}} \end{matrix}$

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