Best solutions to - "Find that solution of y′=2(2x−y) which passes through..."

Carole Yarbrough

Answered question

2021-12-26

Find that solution of

y^{'} = 2 (2 x - y)

which passes through the point (0, 1).

Answer & Explanation

Bukvald5z

Beginner2021-12-27Added 33 answers

Given

y^{'} = 2 (2 x - y) (0, 1)

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

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

= e^{\int p d x} = e^{\int 2 d x} = e^{2 x}

y (I F) = \int (4 x) (I F) d x + c

y (e^{2 x}) = 4 \int x e^{2 x} d x + c

y (e^{2 x}) = 4 [x \int e^{2 x} d x - \int [\frac{d}{d x} (x) \int e^{2 x} d x] d x + c

y (e^{2 x}) = 4 [x \frac{e^{2 x}}{2} - \int \frac{e^{2 x}}{2} d x] + c

y (e^{2 x}) = 4 [\frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4}] + c

y (e^{2 x}) = \frac{4}{4} [2 x e^{2 x} - e^{2 x}] + c

y (e^{2 x}) = [2 x e^{2 x} - e^{2 x}] + c

(a)

y (0) = 1

from (a)

1 (e^{0}) = [2.0 e^{0} - e^{0}] + c

1 = [0 - 1] + c

2 = c

From (a)

y (e^{2 x}) = [2 x e^{2 x} - e^{2 x}] + 2

This is helpful 0 Flag

Durst37

Beginner2021-12-28Added 37 answers

Given: The differential equation

y^{'} = 2 (2 x - y)

and the point P(0,1).
Solution:

y^{'} = 2 (2 x - y)

= 4 x - 2 y

y^{'} + 2 y = 4 x

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

The above differential equation is in the form of first order linear differential equation.

\frac{d y}{d x} + P (x) y = Q (x)

Here

P (x) = 2 and Q (x) = 4 x

Integrating factor

(I F) = e^{\int P (x) d x}

= e^{\int 2 d x}

= e^{2 x}

The general solution is

y \times I F = \int (I F \times Q (x)) d x

y \times e^{2 x} = \int (e^{2 x} \times 4 x) d x

y \times e^{2 x} = 2 e^{2 x} x - e^{2 x} + C

y = \frac{2 e^{2 x} x - e^{2 x} + C}{e^{2 x}}

passes through the point P(0,1)

1 = \frac{2 e^{2 (0)} (0) - e^{2 (0) + C}}{e^{2 (0)}}

1 = \frac{0 - 1 + C}{1}

1 = - 1 + C

C = 2

The solution of the differential equation

y^{'} = 2 (2 x - y)

\Rightarrow y (x) = \frac{2 e^{2 x} x - e^{2 x} + 2}{e^{2 x}}

\Rightarrow y (x) = 2 x - 1 + 2 e^{- 2 x}

user_27qwe

Skilled2022-01-05Added 375 answers

$\begin{matrix} y^{'} = 2 (2 x - y) and the point P(0,1). \\ y^{'} = 2 (2 x - y) \\ = 4 x - 2 y \\ y^{'} + 2 y = 4 x \\ \frac{d y}{d x} + 2 y = 4 x \\ \frac{d y}{d x} + P (x) y = Q (x) \\ Here P (x) = 2 and Q (x) = 4 x \\ (I F) = e^{\int P (x) d x} \\ = e^{\int 2 d x} \\ = e^{2 x} \\ The general solution is \\ y \times I F = \int (I F \times Q (x)) d x \\ y \times e^{2 x} = \int (e^{2 x} \times 4 x) d x \\ y \times e^{2 x} = 2 e^{2 x} x - e^{2 x} + C \\ y = \frac{2 e^{2 x} x - e^{2 x} + C}{e^{2 x}} passes through the point P(0,1) \\ 1 = \frac{2 e^{2 (0)} (0) - e^{2 (0) + C}}{e^{2 (0)}} \\ 1 = \frac{0 - 1 + C}{1} \\ 1 = - 1 + C \\ C = 2 \\ y^{'} = 2 (2 x - y) is \\ \Rightarrow y (x) = \frac{2 e^{2 x} x - e^{2 x} + 2}{e^{2 x}} \\ \Rightarrow y (x) = 2 x - 1 + 2 e^{- 2 x} \end{matrix}$

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