Another yet crazy initial value problem

[Lovely918]

Okay lets say that the initial value problem looked something like this:
2(x+1)y dy/dx + 1+ y^2 =0, y(0)=1
Could you solve it like this:

Like a linear problem:
I was thinking about trying to isolate values of x on one side and y on the other but I'm having problems visualizing it. I don't really know where to start. Should I do this:
2(x+1)y dy/dx =-1(1+y^2)
Then divide and multiply like this
(2(x+1)y dy/dx)/(1+y^2)=-1
(dx)*(2(x+1)y dy/dx)/(1+y^2)=-1 dx
so then id get:
(2(x+1)y dy)/(1+y^2)=-1 dx
Then would I raise it to the e:
e^ln ((2(x+1)y dy)/(1+y^2))=-1x
and then solve for y?

 

[HallsofIvy]

Okay lets say that the initial value problem looked something like this:
2(x+1)y dy/dx + 1+ y^2 =0, y(0)=1
Could you solve it like this:

Like a linear problem:
I was thinking about trying to isolate values of x on one side and y on the other but I'm having problems visualizing it. I don't really know where to start. Should I do this:
2(x+1)y dy/dx =-1(1+y^2)
Then divide and multiply like this
(2(x+1)y dy/dx)/(1+y^2)=-1

Okay, you are good to here. But you cannot leave the "x" on the left side when you are integrating with respect to y. Instead, separate variables by multiplying on both sides by x+ 1.

2y(dy/dx)/(1+ y^2)= -(x+ 1)

[2y/(1+ y^2)]dy= -(x+1)dx
and integrate.

\(\displaystyle (dx)*(2(x+1)y dy/dx)/(1+y^2)=-1 dx
so then id get:
(2(x+1)y dy)/(1+y^2)=-1 dx
Then would I raise it to the e:
e^ln ((2(x+1)y dy)/(1+y^2))=-1x
and then solve for y?\)

\(\displaystyle
No. You are treating that "x" on the left as if it were a constant- and it is not.\)

 

[Lovely918]

Thank you so much!

Thank you for your help

Okay, you are good to here. But you cannot leave the "x" on the left side when you are integrating with respect to y. Instead, separate variables by multiplying on both sides by x+ 1.

2y(dy/dx)/(1+ y^2)= -(x+ 1)

[2y/(1+ y^2)]dy= -(x+1)dx
and integrate.


No. You are treating that "x" on the left as if it were a constant- and it is not.