Solve the following differential equations using the Laplace transform frac{(dy)}{(dt)}-4y=4-2t , y(0)=0

Chesley 2020-12-24 Answered

Solve the following differential equations using the Laplace transform

\frac{(d y)}{(d t)} - 4 y = 4 - 2 t,

y (0) = 0

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Expert Answer

Ayesha Gomez

Answered 2020-12-25 Author has 104 answers

Step 1
Given:

\frac{(d y)}{(d t)} - 4 y = 4 - 2 t

y (0) = 0

\frac{(d y)}{(d t)} - 4 y = 4 - 2 t

Replace

\frac{(d y)}{(d t)}

as y

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New questions

If h has positive derivative and

φ

is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and

φ

is continuous and positive. If:

f (x) = h (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) .

Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:

f^{'} (x) = h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) φ (\frac{x^{4}}{4} - \frac{x^{2}}{2})

We need to analyze where is positive and negative. So I solved the inequalities:

\frac{x^{4}}{4} - \frac{x^{2}}{2} > 0 \land \frac{x^{4}}{4} - \frac{x^{2}}{2} < 0

(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)

for the first case and

(- \sqrt 2, \sqrt 2)

for the second one. Then (not sure of this part)

h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) > 0

φ (\frac{x^{4}}{4} - \frac{x^{2}}{2}) > 0

x \in (- \infty, - \sqrt 2) \cup (\sqrt 2, \infty) .

Also if both h′ and

φ

are negative the product is positive, that's for

x \in (- \sqrt 2, \sqrt 2)

.
The case of the product being negative implies:

x \in [(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)] \cap (- \sqrt 2, \sqrt 2) = [(- \infty, - \sqrt 2) \cap (- \sqrt 2, \sqrt 2)] \cup [(\sqrt 2, \infty) \cap (- \sqrt 2, \sqrt 2)] = \emptyset .

So the function is increasing in

(- \infty, - \sqrt 2), (- \sqrt 2, \sqrt 2), (\sqrt 2, \infty)

. So the function does not have maximum or minimum. Not sure of this but what do you think?

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i need to determine the monotonic intervals of this function

y = 2 x^{3} - 6 x^{2} - 18 x - 7

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My work:

\begin{aligned} y = 2 x^{3} - 6 x^{2} - 18 x - 7 & ⟺ 6 x^{2} - 12 x - 18 = 0 \\ ⟺ 6 (x^{2} - 2 x - 3) = 0 \\ ⟺ (x - 3) (x + 1) \\ ⟺ x - 3 = 0 x + 1 = 0 \\ ⟺ x = 3, x = - 1 \end{aligned}

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and decreases when

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.
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Confusion between the definition of increasing function and a theorem regarding it.
The definition of increasing function given in my school maths text book is
Let I be an open interval contained in the domain of a real valued function f. Then f is said to be increasing on I if

a < b ⟹ f (a) \leq f (b)

a, b \in I

.
And a theorem given after it is
f is increasing in I if

f^{'} (x) > 0 \forall x \in I

.
Shouldn't it be

f^{'} (x) \geq 0

.
Is constant function an increasing function? Or a function like

f (x) = x^{3}

f^{'} (x) = 0

at some or all points which are increasing according to definition.
Edit: The definition of decreasing function given is
f is decreasing on I if

a < b ⟹ f (a) \geq f (b)

.
A constant function follow this definition too.

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Let's say I have $480 to fence in a rectangular garden. The fencing for the north and south sides of the garden costs $10 per foot and the fencing for the east and west sides costs $15 per foot. How can I find the dimensions of the largest possible garden.?

f (x) = \frac{1}{(x - 2) (2 x - 5)}

g (x) = \frac{1}{x^{2}}

, we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at

+ - \sqrt (2 / 5)

1 / \sqrt 2

? 0 because the input is going into g(x) hence as its not in domain of g(x) so it can be said its discontinuous at that point . and similarily other four points as its not in domain of f(g(x)) ? But if suppose we say h(x)=f(g(x)) will we say h(x) is discontinuous at those points (maybe different ones) where x is not defined after simplyfing the whole f(g(x)) to a rational function ?

Proposition: each saddle point is a isolated critical point.
Find a counterexample to disprove it

I am given the function

x (t) = 4 \arctan t

and told that routine computations will show that

x (0) = 0

x (1) = π

.
I must determine a differential equation for

x (t)

x^{'} (t) = f (t, x)

.
I must then use

f

π

.
I will use Euler's Method to find minimum number of iterations (

n

) needed to yield the approximation

π \approx 3.1400

n - 1

iterations is strictly less than 3.1400.
How do I find

n

? It seems to me that it can't be found given the instruction set that I am provided.

Analysis