Differential Equation (1 Viewer)

[skillz]

Hey can someone help me with these.

Find the solution for the following d.e

dy/dx = 9 - y^2, given that y=0 x= 7/6

I did.
dx/dy = 1/(9-y^2)
integrate that becomes x = loge(y+3)/6 - loge(y-3)/6 + c
x-c= 1/6loge ( y+3)/(y-3)
e^6(x-c) = (y+3)/(y-3)

subbing in, y=0 x = 7/6
e^(7/6-c) = -1

now i'm stuck..


2nd q.
dy/dx = root (9- y^2) where x=0 y=3


3rd q

dy/dx = (y^2+2y)/2, given that y=-4, x=0
thanks.

 

[pLuvia]

1.
dy/dx=9-y²
dx/dy=1/(9-y²)
x=int.[dy/(9-y²)]
=ln(y+3)/6-ln(3-y)/6+C
When y=0 x=76 C=76
x=ln(y+3)/6-ln(3-y)/6+76


2.
dy/dx=sqrt{9-y²}
dx/dy=1/sqrt{9-y²}
x=arcsin(y/3)+C
When x=0 y=3 C=-pi/2
x=arcsin(y/3)-pi/2


Is this what you wanted?

 

[pLuvia]

3.
dy/dx=(y²+2y)/2
dx/dy=2/(y²+2y)
x=int.2dy/[y(y+2)]
Using partial fractions
=int.[(1/y)-1/(y+2)]dy
=lny-ln(y+2)+C
When x=0 y=-4 C=?

Are you sure it's a -4?

 

[skillz]

Hey thanks.
however, the answers are different to the book.

for q1. it is.
y= (3(e^(6x-7) -1)/(e^(6x-7)+1)

for q2.

y= 3cos x, -pi<x<0





nfi?

 

[pLuvia]

Ok for my answers just make y the subject and you should get your answers

 

[skillz]

Sorry to sound like a bitch. but i got up to wher you did the x= part
but then i just couldnt change it in to in terms of y

 

[acmilan]

You were correct (almost) in getting to:

x = loge(y+3)/6 - loge(y-3)/6 + c

Remember that int 1/x = ln |x|, the absolute value is important.

x = ln|y+3|/6 - ln|y-3|/6 + c

When y = 0, x = 7/6, so 7/6 = (ln 3)/6 + (ln |-3|)/6 + c -> c = 7/6

6x = ln|y+3| - ln|y-3| + 7

now make y the subject and you'll get what they give.

Remember the absolute value (same for q3)!

pLuvia: check your integration on question 1. Also watch out with your use of QED, its only meant for certain types of questions