Question

If y(x) is the solution of the differential equation
(x+2)dy/dx=x^2+4x-9,x not equal to -2 and y(0),then y(-4) is equal to ?

Answer 1

is it equal to -1?

Answer 2

there is a typo in your question, or an omission. what does y(0) = ?

Answer 3

oops! sorry
y(0)=0
:)

Answer 4

$$
\Large
{
(x+2) \frac{dy}{dx} = x^2 + 4x - 9
\\ \iff
\\ \frac{dy}{dx} = \frac{x^2 + 4x - 9 }{x+2}
\\ \iff
\\ dy = \frac{x^2 + 4x - 9 }{x+2} ~dx
\\ \iff
\\ \int dy = \int \frac{x^2 + 4x - 9 }{x+2}~ dx
\\ \iff \\
y = \frac 12 x^2+2x-13\ln|x+2|+ C
\\
0 = \frac 12~ 0^2+2(0)-13\ln|0+2|+ C
\\ \therefore \\~\\
\\ C = 13 \ln2
\\~\\ \therefore \\
y = \frac 12 x^2+2x-13\ln|x+2|+ 13 \ln 2
\\ \therefore \\
\large {y(-4) = \frac 12 (-4)^2+2(-4)-13\ln|(-4)+2|+ 13 \ln 2
\\= \frac 12 \cdot 16 - 8 - 13 \ln | -2 | + 13\ln 2
\\=8 - 8 - 13 \ln (2) + 13 \ln 2
\\= 0 - 0 = 0
}}



$$

Answer 5

so its equal to 0?

Answer 6

I think so

Answer 7

if we use the absolute values when we integrate. if we don't use the absolute values, then we get an imaginary answer

Answer 8

hhmm..
@rational pls help

Answer 9

\[\dfrac{dy}{dx}=\dfrac{x^2+4x-9}{x+2}\]
\[y(t)-y(0) = \int\limits_0^t\dfrac{x^2+4x-9}{x+2} dx\]

Answer 10

hmm.. what next

Answer 11

are you given the value of \(y(0)\) ?

Answer 12

it must be given, double check the question

Answer 13

yes
i mentioned it later
it's y(0)=0

Answer 14

hmm..

Answer 15

then i think y(-4) is undefined as the domain of solution is (-2, infty)

Answer 16

@Michele_Laino pls help

Answer 17

0,2,1,-1 these are the options

Answer 18

the domain of the solution (-infinity, -2) U (-2, infinity)
the answer is zero, assuming you copied the question correctly

Answer 19

oh okay

Answer 20

note that the absolute value takes care of the case when x < - 2, so we don't have a problem. in general :
$$
\Large \int \frac 1 x dx = \ln | x | + c
$$

Answer 21

I can add a step here

Answer 22

$$\large \rm
y = \begin{cases}
\frac 12 x^2+2x-13\ln(x+2)+ 13 \ln 2 ~if ~x > -2\\
\frac 12 x^2+2x-13\ln(-(x+2))+ 13 \ln 2 ~if~ x < -2
\end{cases}

$$

Answer 23

the inititial value x=0 is on the right side of pole at x=-2 and the value they asked to evaluate is on the left side of pole. So y(-2) does not exist for the solution of this IVP.

Answer 24

http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf

explains clearly about domain of solution of IVP

Answer 25

@rational
"the inititial value x=0 is on the right side of pole at x=-2 and the value they asked to evaluate is on the left side of pole. So y(-2) does not exist for the solution of this IVP."

You mean y(-4) does not exist for the solution of this IVP ?

Answer 26

Ahh yes y(-4)

Answer 27

we could draw a slope field and verify

Answer 28

domain of solution for this IVP is x>-2

Answer 29

i have to read that pdf, but i dont see a problem with it superficially

Answer 30

the attached pdf explains nicely about the domain of solutions and other technical issues

Answer 31

i will do that and get back to you

Answer 32

i cant open the pdf

Answer 33

@IrishBoy123 any comments

Answer 34

domain subtleties

Answer 35

original diff eqn =
\[\frac{dy}{dx} = \frac{(x+2)^2 - 13}{(x+2)}\]
\[= (x + 2) - \frac{13}{x+2}\]

only problem is at x = -2, otherwise looks continuous, the || in the ln
takes care of when x < -2, hence y(-4) = 0

IMHO

Answer 36

okay
thank u

Answer 37

I got this solution:
\[y\left( x \right) = C + \frac{{{x^2}}}{2} + 2x - 13\ln \left| {x + 2} \right|\]

Answer 38

The solution of given IVP is "only" the right part :

\[y = \frac 12 x^2+2x-13\ln(x+2)+ 13 \ln 2 ;~~x\gt -2\]

you cannot have two different solution curves a IVP contradicting the existence and uniqueness thm.

Answer 39

It might seem reasonable to evaluate the integral and simply plugin \(x=-4\). but it wont work here because the domain of IVP is \(x\gt-2\)

Answer 40

@rational

just to be clear, are you saying we know nothing about y when x < -2, other than its derivative?

can we not infer something from the (skewed) symmetry as x+ -> -2 and x- -> -2?

just asking, out of my depth right now.

Answer 41

yes y(-4) can be anything for other IVP.
it is undefined as far as the IVP in the main question is concerned.

Answer 42

the pole at x = -2 and the initial value of (0,0) shrink the domain to (-2, infty)

Answer 43

well out of the four options
0 is the ans right :)

Answer 44

@Haseeb96

Answer 45

yes

Answer 46

the answer is 0?

Answer 47

o par solve kar tu rakha hai ye question .

Answer 48

yes, answer is 0