If tan 0 = 8/15, what is sin 0?
1 answer:
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We will use seperation of variables
tmes both sies by 1/(y^2)
times both sides by dx
we get
1/(y^2)dy=1/(x^3)dx
integrate both sides
-1/y=1/(2x^2)+c
initial condition, (1,1)
-1/1=1/(2(1^2))+c
-1=1/2+c
-3/2=c
-1/y=1/(2x^2)-3/2
solve for y
-1=y(1/(2x^2)-3/2)
y=-1/(1/2x^2-3/2)
tmes both sies by 1/(y^2)
times both sides by dx
we get
1/(y^2)dy=1/(x^3)dx
integrate both sides
-1/y=1/(2x^2)+c
initial condition, (1,1)
-1/1=1/(2(1^2))+c
-1=1/2+c
-3/2=c
-1/y=1/(2x^2)-3/2
solve for y
-1=y(1/(2x^2)-3/2)
y=-1/(1/2x^2-3/2)