Find y if dy/dx = (y-1)/(x+3) and y(-1) = 0
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Given the differential equation:
dy/dx = (y-1)/(x+3)
First we will group the terms of the same variable together.
We will multiply by dx.
==> dy = (y-1)/(x+3) * dx
Now we will divide by (y-1)
==> dy/(y-1) = dx/(x+3)
Now we will integrate both sides.
==> intg (dy/(y-1)) = intg dx/(x+3)
==> ln l y-1 l = ln l x+3l + C
Let us rewrite using the exponent form.
==> e^(ln l y-1l) = e^(ln lx+3l + C)
==> l y -1 l = e^ln l x+3l * e^C
==> l y-1 l = l x+ 3l * C1
==> l y-1 l = C1 * l x+3l
==> y-1 = +- C1 (x+3)
==> y = 1 + - C1 ( x+3)
Now we are given that y(-1) = 0
==> 0 = 1 +- C1 ( -1+3)
==> 0 = 1 +- C1 * 2
==> 0 = 1+-2C1
==> C1 = +-1/2
==> y = 1 +- (1/2) (x+3)
==> y= 1+- (x+3)/2
==> y = (2+(x+3)/2 = (x+5)/2
==> y= (2-x-3)/2 = -(x+1)/2
Then we have two possible values for y.
y1= (x+5)/2
y2- -(x+1)/2
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