Write an equivalent first-order differential equation and initial condition for y. y=−4+∫1x(4t−y(t))dt What is the equivalent first-order differential equation? What is the initial condition?

Name: Write an equivalent first-order differential equation and initial condition for y. y=−4+∫1x(4t−y(t))dt What is the equivalent first-order differential equation? What is the initial condition?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Write an equivalent first-order differential equation and initial condition for y. y=−4+∫1x(4t−y(t))dt What is the equivalent first-order differential equation? What is the initial condition?

y=−4+∫1x(4t−y(t))dt Differentiating both side, we get dydx=ddx[∫1x(4t−y(t))dt] or, ddxy(x)=4x−y(x) ∴ dydx+y=4x → equivalent at order differential equation Initial condition is at x=1 ∴ y(1)=−4+∫11(4t−y(t))dt y(1)=−4 [∵ ∫aaf(x)dx=0] ∴ at{x=1y=−4 → initial condition

y=−4+∫1x(4t−y(t))dtdydx=0+ddx(∫1x(et−y(t))dtddx∫h(x)g(x)f(t)dt=f(h(x))h′(x)−f(g(x))g′(x)=[4(1)−y(1)]0−[4x−y(x)]×1dydx=y(x)−4xy′−y+4x=0⇒ x=1 y(1)=−4

y=−4+∫1x(4t−y(t))dt Differentiating both side, we get dydx=ddx[∫1x(4t−y(t))dt] or, ddxy(x)=4x−y(x) ∴ dydx+y=4x → equivalent at order differential equation Initial condition is at x=1 ∴ y(1)=−4+∫11(4t−y(t))dt y(1)=−4 [∵ ∫aaf(x)dx=0] ∴ at{x=1y=−4 → initial condition

y=−4+∫1x(4t−y(t))dtdydx=0+ddx(∫1x(et−y(t))dtddx∫h(x)g(x)f(t)dt=f(h(x))h′(x)−f(g(x))g′(x)=[4(1)−y(1)]0−[4x−y(x)]×1dydx=y(x)−4xy′−y+4x=0⇒ x=1 y(1)=−4

Expert answers to "Write an equivalent first-order differential equation and initial..."

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Calculus and Analysis
Differential Equations
First order differential equations

Kaspaueru2

Answered question

2022-01-22

Write an equivalent first-order differential equation and initial condition for y.

y = - 4 + \int_{1}^{x} (4 t - y (t)) d t

What is the equivalent first-order differential equation?
What is the initial condition?

Answer & Explanation

Esta Hurtado

Beginner2022-01-22Added 39 answers

$y = - 4 + \int_{1}^{x} (4 t - y (t)) d t$
Differentiating both side, we get
$\frac{d y}{d x} = \frac{d}{d x} [\int_{1}^{x} (4 t - y (t)) d t]$
or, $\frac{d}{d x} y (x) = 4 x - y (x)$
$∴ \frac{d y}{d x} + y = 4 x \to$ equivalent at order differential equation
Initial condition is at x=1
$∴ y (1) = - 4 + \int_{1}^{1} (4 t - y (t)) d t$
$y (1) = - 4 [∵ \int_{a}^{a} f (x) d x = 0]$
$∴ a t {\begin{matrix} x = 1 \\ y = - 4 \end{matrix} \to$ initial condition

Linda Birchfield

Beginner2022-01-23Added 39 answers

y = - 4 + \int_{1}^{x} (4 t - y (t)) d t

\frac{d y}{d x} = 0 + \frac{d}{d x} (\int_{1}^{x} (e t - y (t)) d t

\frac{d}{d x} \int_{h (x)}^{g (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)

= [4 (1) - y (1)] 0 - [4 x - y (x)] \times 1

\frac{d y}{d x} = y (x) - 4 x

y^{'} - y + 4 x = 0

\Rightarrow x = 1 y (1) = - 4

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