Question

can someone please help me with solving differential equations?

Answer 1

@ganeshie8 @phi can you please help me when you get a chance?!

Answer 2

can you "separate the variables"?

Answer 3

im doing really bad with solving them, and i dont really understand what my teacher is saying

Answer 4

like (y/1)(cosx/y^2)?

Answer 5

\[ \frac{dy}{dx}= \frac{y \cos x }{1+y^2} \]
multiply both sides by (1+y^2)/y
what do you get ?

Answer 6

or do it in steps.
first multiply both sides by 1/y

Answer 7

then multiply both sides by (1+y^2)

Answer 8

if im following this right, it should be 1+y^2/y=cosx?

Answer 9

except you don't leave out dy/dx
you get
\[ \frac{dy}{dx}= \frac{y \cos x }{1+y^2} \\ \frac{(1+y^2)}{y}\frac{dy}{dx}= \cos x \]
now multiply both sides by dx

Answer 10

the idea is you get dy and *all* the y variables on one side
ditto for x and dx on the other side

Answer 11

ohh wow i did that by accident, totally forgot about it haha

Answer 12

after multiplying you get ?

Answer 13

(1+y^2)/ydy=cosxdx?

Answer 14

yes.
\[ \frac{(1+y^2)}{y}\ dy= \cos x \ dx\]
we could divide the y into each term up top, and re-write it as
\[ \frac{1}{y} + y \ dy = \cos x \ dx \]
and though we generally don't write in the parens the left side is really
\[ \left( \frac{1}{y} + y\right) \ dy = \cos x \ dx \]

Answer 15

now integrate
\[ \int \left( \frac{1}{y} + y\right) \ dy = \int \cos x \ dx \]

Answer 16

for the left side, integrate term by term.
for the right side, you have to know your integral of cos (memorize)

Answer 17

can you do the integration ?

Answer 18

yess, ill do it now

Answer 19

don't forget the constant of integration.

Answer 20

ln(y)+y^2/2+C for the first side? @phi @ganeshie8 @hartnn

Answer 21

and the second side sin(x)+C @ganeshie8 @hartnn can you please chck since phi had to go?

Answer 22

@phi

Answer 23

yes, but you only have one constant. (If you have C1 on the left side, and C2 on the right side) you can combine them to get one constant

Answer 24

except for the constant C, i accidently addded it

Answer 25

oh hahahah

Answer 26

now replace x with 0 and y with 1 and solve for C

Answer 27

C= ln+1?

Answer 28

ln(y)+y^2/2+C = sin x

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

Answer 29

you should know the values for ln(1) and sin(0)
but a calculator will tell you. (But I would learn these values)

Answer 30

ohh i did it over 1 by accident

Answer 31

C=-1/2

Answer 32

yes. so the final answer is
\[ \ln y + \frac{1}{2} y^2 - \frac{1}{2} = \sin x \]

Answer 33

or some variation, as we can write it in a few different (equivalent) ways.

Answer 34

thank you(: