ODE question... (1 Viewer)

[Zarathustra]

Hey,
Would anyone be kind enough to post a solution to this ODE?

2*e^x+(dy/dx)*(1-e^x)*tan(y)=0


The answer given is sec(y)=C(e^x-1)^2

 

[hyparzero]

2e^x+(dy/dx)*(1-e^x)tan(y)=0

Rearraging like Iruka said:

(dy/dx)*(1-e^x)tan(y) = - 2e^x

Hence

(dy/dx)*tan(y) = [ 2e^x ] / [ e^x - 1 ]

∫tan(y)dy = ∫[ 2e^x ] / [ e^x - 1 ] dx

=> - ln[cos(y)] = 2ln[ e^x - 1 ] + C ............. Take exp of both sides..

=> 1 / cos(y) = C(e^x - 1)²

=> sec(y) = C(e^x - 1)²

 

[Zarathustra]

Thanks guys... much appreciated.

 

[STx]

∫tan(y)dy = ∫[ 2e^x ] / [ e^x - 1 ] dx

=> - ln[cos(y)] = 2ln[ e^x - 1 ] + C ............. Take exp of both sides..

=> 1 / cos(y) = C(e^x - 1)²

=> sec(y) = C(e^x - 1)²

After 'Take exp of both sides', why does the C multiply the (e^x - 1) ?

 

[Slidey]

Exponentiation:
a^(b+c)=a^c.b^c

In this case it's like, for a constant, x variable:
e^(x+a)=e^a.e^x, e^a is a constant, so it equals Ce^x. Whenever you change the value of a constant it's a good idea to change the letter. If you used C before, use K now. Hyparzero didn't do this.

 

[hyparzero]

Exponentiation:
a^(b+c)=a^c.b^c

In this case it's like, for a constant, x variable:
e^(x+a)=e^a.e^x, e^a is a constant, so it equals Ce^x. Whenever you change the value of a constant it's a good idea to change the letter. If you used C before, use K now. Hyparzero didn't do this.

i think you mean:
a^(b+c)=a^b * a^c

not

a^(b+c)=a^c.b^c


yea, i'm always used to using the same constant letter throughout, saves me alot of confusion and time ~ as if you introduce a new constant, you'll have to state, again; "where [ ] is a constant"

would i get marked down for this or something?

 

[Zarathustra]

Ahh... could someone (I'm assuming hyparzero) write the solution to this ODE up as well??? Yes, I suck at this stuff....
2*x*e^y/x+y-x*(dy/dx)=0

P.S. Slide rule, I thought you were doing plant science??????

 

[Zarathustra]

Thanks again Iruka