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Integration problem...

dy/dx = -6sin(x)y INITIAL CONDITIONS: y(0)=-2

So i did this:

dy/y = -6sin(x)dx integrate both sides

ln(y) = 6cos(x) + c

Now tricky conditions as you can't ln a negative number:

y = exp(6cos(x) + c)

I'm stuck from here, but the answer given is y = -2(x)exp[6cos(x)-6]

Can anyone help here?
Reply 1
Avatar for StewieP
StewieP
OP
anyone able to help?
Reply 2
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StewieP
OP
Is no help going to come my way?!
Reply 3
Avatar for JXamie
JXamie
12
Original post by StewieP
dy/dx = -6sin(x)y INITIAL CONDITIONS: y(0)=-2

So i did this:

dy/y = -6sin(x)dx integrate both sides

ln(y) = 6cos(x) + c

Now tricky conditions as you can't ln a negative number:

y = exp(6cos(x) + c)

I'm stuck from here, but the answer given is y = -2(x)exp[6cos(x)-6]

Can anyone help here?


You've done it fine I think.

At the bolded point I would rewrite this as:

y=Ae6cosxy=Ae^{6cosx}

(As ece^c can be written as just another constant).

Inputting the initial conditions you get A=−2A=-2 and therefore:

y=−2e6cosxy=-2e^{6cosx}

Are you sure the answer you have is correct? :confused:
Reply 4
Avatar for StewieP
StewieP
OP
Original post by JXamie
You've done it fine I think.

At the bolded point I would rewrite this as:

y=Ae6cosxy=Ae^{6cosx}

(As ece^c can be written as just another constant).

Inputting the initial conditions you get A=−2A=-2 and therefore:

y=−2e6cosxy=-2e^{6cosx}

Are you sure the answer you have is correct? :confused:


You don't though?

You get -2 = Ae^6

That doesn't give A=2
Reply 5
Avatar for Inactivity
Inactivity
Original post by JXamie
You've done it fine I think.

At the bolded point I would rewrite this as:

y=Ae6cosxy=Ae^{6cosx}

(As ece^c can be written as just another constant).

Inputting the initial conditions you get A=−2A=-2 and therefore:

y=−2e6cosxy=-2e^{6cosx}

Are you sure the answer you have is correct? :confused:


No, cos(0) = 1, so A=−2e−6A=-2e^{-6} therefore y=−2e−6e6cos(x),y=−2e6cos(x)−6y=-2e^{-6}e^{6cos(x)}, y=-2e^{6cos(x) -6}
Reply 6
Avatar for StewieP
StewieP
OP
Original post by Inactivity
No, cos(0) = 1, so A=−2e−6A=-2e^{-6} therefore y=−2e−6e6cos(x),y=−2e6cos(x)−6y=-2e^{-6}e^{6cos(x)}, y=-2e^{6cos(x) -6}


Thanks!
Reply 7
Avatar for JXamie
JXamie
12
Original post by Inactivity
No, cos(0) = 1, so A=−2e−6A=-2e^{-6} therefore y=−2e−6e6cos(x),y=−2e6cos(x)−6y=-2e^{-6}e^{6cos(x)}, y=-2e^{6cos(x) -6}


Ah.. basic maths knowledge fail :tongue: Cheers :smile:

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