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Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answ

Ivan

Answer:

$(-\infty,3)$

Step-by-step explanation:

We are given that

$(x-3)y''+4y=x$

$y''+\frac{4}{x-3}y=\frac{x}{x-3}$

y(0)=0

y'(0)=1

By comparing with

$y''+p(x)y'+q(x)y=g(x)$

We get

$p(x)=\frac{4}{x-3}$

$g(x)=\frac{x}{x-3}$

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

Interval on which p(x),q(x) and g(x) are continuous

$(-\infty,3)$ and (3, $\infty)$

By unique existence theorem

Largest interval which contains 0= $(-\infty,3)$

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution= $(-\infty,3)$

7 0

3 years ago

What 228,643 to the nearest hundred

Annette [7]

228,643 to the nearest hundered is 228,600.

3 0

4 years ago

I need help please, I am so confused.

vazorg [7]

Answer:

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

Step-by-step explanation:

x-intercepts are when the curve intercepts the x-axis, so when y =0.

Therefore, to find the x-intercepts, substitute y = 0 and solve for x.

The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.

Equation: $y=x(x-2)$

$\implies x(x-2)=0$

$\implies x=0$

$\implies (x-2)=0 \implies x=2$

Therefore, the x-intercepts are x = 0 and x = 2

The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1

Equation: $y=(x-4)(x+5)$

$\implies (x-4)(x+5)=0$

$\implies (x-4)=0 \implies x=4$

$\implies (x+5)=0 \implies x=-5$

Therefore, the x-intercepts are x = -5 and x = 4

The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5

Equation: $y=-5x(3-x)$

$\implies -5x(3-x)=0$

$\implies -5x=0 \implies x=0$

$\implies (3-x)=0 \implies x=3$

Therefore, the x-intercepts are x = 0 and x = 3

The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

4 0

2 years ago

How do I solve for x and what equation do I use to solve

Yuri [45]

You didnt show the rest of the picture i can't help if i can't see.

8 0

3 years ago

Eighteen students decide they don't want to be

Ratling [72]

Answer: 45

Step-by-step explanation:

18 = 40% N

18 = 0.4 N

Divide both sides by 0.4

N = 45

6 0

3 years ago

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