Let y=y(x) be in the solution of the differential equation cosx(3sinx+cosx+3) dy=(1+y sinx(3sinx+cosx+3))dx, 0 ≤ x ≤ π/2, y(0) = 0. Then y(π/3) is equal to:
Let y=y(x) be in the solution of the differential equation cosx(3sinx+cosx+3) dy=(1+y sinx(3sinx+cosx+3))dx, 0 ≤ x ≤ π/2, y(0) = 0. Then y(π/3) is equal to:
Option 1 -
2logₑ((2√3 + 10)/11)
Option 2 -
2logₑ((2√3 + 9)/6)
Option 3 -
2logₑ((√3 + 7)/2)
Option 4 -
2logₑ((3√3 - 8)/4)
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1 Answer
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Correct Option - 1
Detailed Solution:cos(x)(3sin(x) + cos(x) + 3)dy = (1 + ysin(x)(3sin(x) + cos(x) + 3))dx
This seems mistyped. A more likely form is:
dy/dx - (sin(x)/(cos(x)))y = 1 / (cos(x)(3sin(x) + cos(x) + 3))
dy/dx - tan(x)y = sec(x) / (3sin(x) + cos(x) + 3)The integrating factor (I.F.) is:
I.F. = e^∫(-tan(x))dx = e^(ln|cos(x)|) = cos(x).Multiplying by I.F.:
d(y*cos(x))/dx = 1 / (3sin(x) + cos(x) + 3)y*cos(x) = ∫ dx / (3sin(x) + cos(x) + 3)
Using Weierstrass substitution, let t = tan(x/2):
sin(x) = 2t/(1+t²), cos(x) = (1-t²)/(1+t²), dx = 2dt/(1+t²)∫ (2dt/(1+t²)) / (3(2t/(1+t²)) + (1-t²)/(1+t²) + 3)
= ∫ 2dt / (6t + 1 -...more
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