Kaspaueru2

2022-01-22

Write an equivalent first-order differential equation and initial condition for y.
$y=-4+{\int }_{1}^{x}\left(4t-y\left(t\right)\right)dt$
What is the equivalent first-order differential equation?
What is the initial condition?

$y=-4+{\int }_{1}^{x}\left(4t-y\left(t\right)\right)dt$
Differentiating both side, we get
$\frac{dy}{dx}=\frac{d}{dx}\left[{\int }_{1}^{x}\left(4t-y\left(t\right)\right)dt\right]$
or, $\frac{d}{dx}y\left(x\right)=4x-y\left(x\right)$
equivalent at order differential equation
Initial condition is at x=1

initial condition

Linda Birchfield

$y=-4+{\int }_{1}^{x}\left(4t-y\left(t\right)\right)dt$
$\frac{dy}{dx}=0+\frac{d}{dx}\left({\int }_{1}^{x}\left(et-y\left(t\right)\right)dt$
$\frac{d}{dx}{\int }_{h\left(x\right)}^{g\left(x\right)}f\left(t\right)dt=f\left(h\left(x\right)\right){h}^{\prime }\left(x\right)-f\left(g\left(x\right)\right){g}^{\prime }\left(x\right)$
$=\left[4\left(1\right)-y\left(1\right)\right]0-\left[4x-y\left(x\right)\right]×1$
$\frac{dy}{dx}=y\left(x\right)-4x$
${y}^{\prime }-y+4x=0$

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