Question

Please help! Will FAN AND MEDAL!

Solve the differential equation dy dx equals the quotient of y times the cosine of x and the quantity 1 plus y squared with the initial condition y(0) = 1.

Answer 1

Can you please help? @mathmale @Nnesha @Directrix

Answer 2

\[\large\rm \frac{ dy }{ dx }=\frac{ ycos(x) }{ (1+y^2) }\] like this or
\[\large\rm \frac{ dy }{ dx }=\frac{ ycos(x) }{ (1+y)^2 }\]

Answer 3

?

Answer 4

hi ??

Answer 5

well i gtg so i think 2nd one is correct
first isolate dy
\[\large\rm 1dy=\frac{ ycos(x) }{ (1+y)^2 } dx\]
and then integrate \[\large\rm \int\limits_{}^{}1dy=\int\limits_{}^{}\frac{ ycos(x) }{ (1+y)^2 } dx\] find the anti-derivative of both sides

y(0)=1 represents that when x=0 y=1
so at the end substitute x and y for these numbers

Answer 6

If other users have to ask you for clarification about what you meant by a verbal descrption of a math problem situation, that's a clue that you should perhaps be honing your presentation skills using math symbols instead of words.

Answer 7

I am so sorry I had to help my dad with something. The first one is correct. @mathmale can you please help me?

Answer 8

separate the variables and integrate
\[\int\limits \frac{ 1+y^2 }{ y }dy=\int\limits \cos x dx+c\]

Answer 9

Ohh okay is that is?

Answer 10

@surjithayer

Answer 11

@surjithayer

Answer 12

\[\int\limits \frac{ 1 }{ y }dy+\int\limits y~dy=\sin x+c\]
\[\ln y+\frac{ y^2 }{ 2 }=\sin x+c\]
when x=0,y=1
\[\ln 1=0\]

Answer 13

find c and then substitute the value of c

Answer 14

C=-sin(x)+ln(y)+y^2/2 is that correct? @surjithayer

Answer 15

Am I right @Nnesha

Answer 16

that's where you need to use the given info y(0)=1
when x=0 y=1
substitute x for 0 y for 1

Answer 17

okay but what part of the equation do I use. there are so many steps I am not 100% sure where to plug in 0 and 1

Answer 18

you're good to plugin 0 and 1 right after you find out the anti derivative

Answer 19

what do I find the anti derivative of?

Answer 20

@Nnesha

Answer 21

hmm look at surjithayer's comment that's a good way to start

Answer 22

to get rid of the fraction and multiplication he moved (y/1+y^2) to the left side

Answer 23

C=1/2?

Answer 24

how did you get 1/2 ?

Answer 25

\(\color{blue}{\text{Originally Posted by}}\) @aryana_maria2323
C=-sin(x)+ln(y)+y^2/2 is that correct?
\(\color{blue}{\text{End of Quote}}\)
substitute x for 0 y=1

Answer 26

ln(1)+1^2/2=sin(0)+C
C=1/2

Answer 27

ye.

Answer 28

So im correct?

Answer 29

yes

Answer 30

Yay thank you so much.