Question

find the particular solution for \[xyy' = 1 + y^2\] when x = 2; y = 3

the book says it is \(y = \frac{1}{2} \sqrt{ 10x^2 - 4}\) and idk how it got that

Answer 1

i got \[\large \frac{1}{2} \ln (1 + y^2) = \ln x + \frac{\sqrt{10}}{2}\]

Answer 2

\[ \int \frac{y}{1 + y^2} dy = \int \frac{dx}{x}\]
\[ \ln \sqrt{1+y^2} = \ln x + \ln C\]
\[ \sqrt{1+ y^2} = xC\]
\[ y = +- \sqrt{Cx^2 - 1} \]

How do you write +-

Answer 3

\pm

Answer 4

find the value of C .... using initial values given!!
\[ y = \pm \sqrt{Cx^2 - 1} \]

Answer 5

so it would be \[\large y = \pm \sqrt{\frac{\sqrt{10}}{2} x^2 - 1}\] how did that become 10x^2 - 4?

Answer 6

btw..do i always express it as y equals?

Answer 7

C = 10/4 <--- you don't need to take root

Answer 8

how did you get 10/4?

Answer 9

also in your answer you should be getting
\[ C = \ln \frac{\sqrt{10}}{2}\]
cancelling and isolating y you should be able to get the same

Answer 10

\[\large \sqrt{1+y^2} = Cx\] \[\sqrt{1+ (3)^2} = c(2)\]
\[\sqrt{10} = 2C\]

Answer 11

C <-- integral constants are stupid ... try taking constants such that i helps to simplify your equation or expression

so C = sqrt{10} by 2
just simplify that equation in terms of y,you will get the same ...

Answer 12

how do i simplify in terms of y?

Answer 13

uhmm okay i got it lol

Answer 14

C^2 is still C

Answer 15

anyway how do you turn \[\sqrt{\frac{10}{4} x^2 - 1}\] into 10x^2 - 4? no denominators?

Answer 16

\[ \sqrt{1 + y^2} = \frac{\sqrt{10}}{2}x\]
square both sides
\[ 1+y^2 = \frac{10x^2}{4}\]
\[y^2 = \frac{10x^2 - 4}{4} \]
take square root ...

Answer 17

oh nevermind i saw it lol...just one question...do i always put it in terms of y equals?

Answer 18

not necessarily ... there might be some equation whose form cannot be expressed simply interms of y ... leave those.

and besides there are not functions ... just solution curves

Answer 19

*these <-- there

Answer 20

i see thanks

Answer 21

yw