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Can anyone pls help with this math question, all steps would be appreciated:
(dy/dx)=(y/x-3), where x can't equal 3. Find the particular solution to y=f(x) with the initial condition f(-1)=1
The answer is y=(-1/4)x+(3/4)
pls help, will luv u
muchas gracias mes amis
please @ me if you can help, I need to study so I won't be able to check frequently
Last edited by Aura; 06-01-2017 at 08:32 PM.
i love luna
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i love luna
good luck! i can't math at all :c
Aura (06-01-2017)
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It's a separable differential equation.
Just came back from work and kind of tired but:
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Just keep in mind how to do the constant part though I think those websites will show you.
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Aura (06-01-2017)
Okay, its a separable equation.
First, you know x can't equal 3 because the denominator can't be 0. So if x-3=0, x=3, therefore x=/=3
So, you multiply both sides by dx and divide both sides by y to get
dy/y=dx/(x-3)
Then you integrate both sides.
integration of 1/y is ln|y|+C
integration of 1/(x-3) is ln|x-3|+C
So you get
ln|y|=ln|x-3|+C1
using the property of logarithms
ln|y|=ln|e^C1(x-3)|
multiply by e both sides
y=e^c1(x-3) (since e^c1 is constant)
y=c(x-3)
sub in initial conditions y=1, x=-1
1=c((-1)-3)
1=c(-4)
-1/4=c
plug in
y=(-1/4)(x-3)
y=-1/4x+3/4
therefore your final equation is y=-1/4x+3/4 where x=/=3
Last edited by kat.; 06-02-2017 at 07:40 AM.
Aura (06-02-2017)