Answer:
0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.
Step-by-step explanation:
Given the differential equation
(t - 5)y' + (ln t)y = 6t
and the condition y(1) = 6
We can rewrite the differential equation by dividing it by (t - 5) as
y' + [(ln t)/(t - 5)]y = 6t/(t - 5)
(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).
6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)
We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).
But the initial condition is y = 6, when t = 1.
The solution to differential equation is certain to exist at (0, 5)
Which implies that
0 < t < 5
is the required interval.
Step-by-step explanation:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
56 = 2 ×2 × 2 × 7
Now
Common factor = 2
Remaining factor = 2 × 3 × 2 × 3 × 7
LCM = RF × CF
= 504
hence the lCM of 12 , 18 and 56 is 504...
<span><span>Graph <span>x2<span> = 4</span>y</span><span> and state the vertex, focus, axis of symmetry, and directrix.</span></span><span>This is the same graphing that I've done in the past: </span><span>y = (1/4)x2</span><span>. So I'll do the graph as usual:</span></span><span> </span><span>The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:<span>x2 = 4y</span> Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved<span>
(x – 0)2 = 4(y – 0)</span><span>This rearrangement "shows" that the vertex is at </span><span>(h, k) = (0, 0)</span><span>. The axis of symmetry is the vertical line right through the vertex: </span><span>x = 0</span>. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.<span>From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that </span><span>4p = 4</span><span>, so </span><span>p = 1</span><span>. Then the focus is one unit above the vertex, at </span>(0, 1)<span>, and the directrix is the horizontal line </span><span>y = –1</span>, one unit below the vertex.<span>vertex: </span>(0, 0)<span>; focus: </span>(0, 1)<span>; axis of symmetry: </span><span>x<span> = 0</span></span><span>; directrix: </span><span>y<span> = –1</span></span></span><span><span><span>Graph </span><span>y2<span> + 10</span>y<span> + </span>x<span> + 25 = 0</span></span>, and state the vertex, focus, axis of symmetry, and directrix.</span><span>Since the </span>y<span> is squared in this equation, rather than the </span>x<span>, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking </span>y<span>-values first and then finding the corresponding </span>x<span>-values for </span><span>x = –y2 – 10y – 25</span>:<span>To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:</span><span>y2 + 10y + 25 = –x</span> <span>
(y + 5)2 = –1(x – 0)</span><span>This tells me that </span><span>4p = –1</span><span>, so </span><span>p = –1/4</span><span>. Since the parabola opens to the left, then the focus is </span>1/4<span> units to the left of the vertex. I can see from the equation above that the vertex is at </span><span>(h, k) = (0, –5)</span><span>, so then the focus must be at </span>(–1/4, –5)<span>. The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is </span>1/4<span> units to the right of the vertex. Putting this all together, I get:</span><span>vertex: </span>(0, –5)<span>; focus: </span>(–1/4, –5)<span>; axis of symmetry: </span><span>y<span> = –5</span></span><span>; directrix: </span><span>x<span> = 1/4</span></span></span><span><span>Find the vertex and focus of </span><span>y2<span> + 6</span>y<span> + 12</span>x<span> – 15 = 0</span></span></span><span><span>The </span>y<span> part is squared, so this is a sideways parabola. I'll get the </span>y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.<span>y2 + 6y – 15 = –12x</span> <span><span>
y</span>2 + 6y + 9 – 15 = –12x + 9</span> <span>
(y + 3)2 – 15 = –12x + 9</span> <span>
(y + 3)2 = –12x + 9 + 15 = –12x + 24</span> <span>
(y + 3)2 = –12(x – 2)</span> <span>
(y – (–3))2 = 4(–3)(x – 2)</span></span><span><span>Then the vertex is at </span><span>(h, k) = (2, –3)</span><span> and the value of </span>p<span> is </span>–3<span>. Since </span>y<span> is squared and </span>p<span> is negative, then this is a sideways parabola that opens to the left. This puts the focus </span>3 units to the left of the vertex.<span>vertex: </span>(2, –3)<span>; focus: </span><span>(–1, –3)</span><span>
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