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Mathematics 14 Online
OpenStudy (anonymous):

Find the domain for the particular solution to the differential equation dy dx equals the quotient of negative 1 times x and y, with initial condition y(2) = 2.

OpenStudy (anonymous):

https://gyazo.com/6439d6ba1afb5baf216c9b05cf9296f0

OpenStudy (anonymous):

@timo86m would you know?

OpenStudy (anonymous):

sorry haven't taken DE yet

OpenStudy (anonymous):

It's cool. I appreciate you taking your time to read the question.

OpenStudy (anonymous):

@freckles Would you happen to know?

OpenStudy (freckles):

first find when when dy/dx does not exist note dy/dx is -x/y and -x/y is fine unless ....?

OpenStudy (freckles):

think fractions are not cool if the bottom is ...

OpenStudy (anonymous):

0

OpenStudy (freckles):

right we don't want y to be 0... now we need to keep this in mind after we solve the differential equation we will try to use this so what do you get when you solve dy/dx=-x/y

OpenStudy (anonymous):

I got \[y = \sqrt{\frac{ -x^2 }{ 4 }+5}\]

OpenStudy (freckles):

not sure how you got the 5... \[y dy=-x dx \\ \frac{y^2}{2}=\frac{-x^2}{2}+C \\ \text{ I would \not attempt \to solve for } y \text{ yet } \\ \text{ but I'm going \to use the condition given \to find } C \\ \frac{2^2}{2}=\frac{-2^2}{2}+C \\ \frac{2(2)^2}{2}=C \\ C=2^2=4 \\ \frac{y^2}{2}=\frac{-x^2}{2}+4 \\ y^2=-x^2+8 \\ \text{ but } y \neq 0 \\ \text{ so } -x^2+8 \neq 0 \text{ when ? }\]

OpenStudy (anonymous):

What are you asking exactly?

OpenStudy (freckles):

for what x is -x^2+8=0?

OpenStudy (anonymous):

\[\sqrt{8}\]

OpenStudy (freckles):

or?

OpenStudy (anonymous):

-\[\sqrt{8}\]

OpenStudy (freckles):

|dw:1449725022181:dw|

OpenStudy (anonymous):

-2sqrt(2) to 2sqrt(2)

OpenStudy (freckles):

right

OpenStudy (anonymous):

Thanks

OpenStudy (anonymous):

Just to confirm

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