You use additive inverse to combine like terms that are on opposite sides of the equation.
ie.
3x + 1 = 6x -12
You use additive inverse to move -12 to the +1

3 0

3 years ago

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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

1 1/9 • 6= x what is x

brilliants [131]

This problem basically already gives you X, you just have to simplify.

To complete this problem all you have to do is divide 11 by 9.6

11 / 9.6 = 1.15

So X = 1.15

5 0

3 years ago

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What's the next answer in BADEHGJK

alina1380 [7]

BADEHGJK NM because it's going backwards and then it skips a letter

3 0

3 years ago