Thread: Seperation of variables.

4)general sol: (dv/dt)=(4(v^2)+36)(t^2)

Originally Posted by kmankish

1)solve: (2x+7t)(dx/dt)+7x-t=0, x(0)=5, x(t)=?
2)general sol: (dx/dt)+5x=(x^2)(e^-2t)
3)solve: (dy/dx)+(y/x)=(y^6)(x^5), y(1)=1, y(x)=?
4)general sol: (dv/dt)=(4(v^2)+36)(t^2)

No, that's not how this site works. You need to show effort, and post one question per thread.

yh i no ive been working on these q's for ages now and i keep running into dead ends i was hoping someone would help me with the method for atleat 1 which ever they prefer so i can find out how to do it.

Originally Posted by kmankish

yh i no ive been working on these q's for ages now and i keep running into dead ends i was hoping someone would help me with the method for atleat 1 which ever they prefer so i can find out how to do it.

I understand, but try to see this from a different point of view:
There are people on MHF who are trying to offer *help* (that's what the "H" stands for in MHF). If you want help, I don't think it's unreasonable that you be asked to show some work. Furthermore, this most recent post is very off-putting. It may be that English isn't your native language, but you might still try to clean it up a bit!

Finally, here's a site that will help you format your questions in a pretty way. Online LaTeX Equation Editor - create, integrate and download

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Now, #4 looks like it's "separable"

We get

Can you integrate both sides? A trig substitution might help on the left...

yes i understand, thanks for the suggestion am i right in thinking the rhs goes to

if i have c on both sides can i take it as c overall?

Originally Posted by kmankish

yes i understand, thanks for the suggestion am i right in thinking the rhs goes to

if i have c on both sides can i take it as c overall?

Looks good.
Now say you have C_1 on the left and C_2 on the right. Then you can subtract C_1 from both sides and let "C" = C_2 - C_1