Find the equation of the curve represented by dy/dx+ycotx=5e^cosx passing through the point (pi/2,2)
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Given ODE of first degree:
y' + Cotx y = 5 e^(cos x)
Looking at RHS, it is useful to multiply both sides by Sin x.
Sin x y' + cos x y = 5 Sin x * e^(cos x)
d(y Sin x)/dx = 5 * d[e^(cos x)]/dx
Integrate both sides: y Sin x = 5 e^(cos x) + C
y = 5 Cosec x * e^(Cos x) + C Cosec x
The above curve passes through (π/2, 2). So
2 = 5 * 1 * e^0 + C
=> C = -3
The curve: y = Cosec x * [ 5 e^(cos x) - 3]
y' + Cotx y = 5 e^(cos x)
Looking at RHS, it is useful to multiply both sides by Sin x.
Sin x y' + cos x y = 5 Sin x * e^(cos x)
d(y Sin x)/dx = 5 * d[e^(cos x)]/dx
Integrate both sides: y Sin x = 5 e^(cos x) + C
y = 5 Cosec x * e^(Cos x) + C Cosec x
The above curve passes through (π/2, 2). So
2 = 5 * 1 * e^0 + C
=> C = -3
The curve: y = Cosec x * [ 5 e^(cos x) - 3]
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