How do we integrate the following with respect to x? Note that y is a function of x
(dy/dx)/(((y^2) - c)^(1/2)) = +/-(1/(sqrt(c))
To integrate the given expression with respect to x, we can use the technique of separation of variables. Firstly, we can multiply both sides by ((y^2) - c)^(1/2) dx to separate the variables y and x as follows:
(dy/((y^2) - c)^(1/2))) = +/-(dx/(sqrt(c)))
Now, we can integrate both sides with respect to their respective variables as follows:
∫(dy/((y^2) - c)^(1/2))) = +/-(∫(dx/(sqrt(c))))
Using the substitution u = y^2 - c, du/dy = 2y, and du = 2y dy, we can simplify the left-hand side integral as follows:
∫(dy/((y^2) - c)^(1/2))) = ∫(1/((y^2) - c)^(1/2)) (du/(2y)) = (1/2) ln|u| + C1
where C1 is the constant of integration.
Similarly, we can simplify the right-hand side integral using the substitution v = (1/sqrt(c)) x, dv/dx = (1/sqrt(c)), and dx = sqrt(c) dv, as follows:
∫(dx/(sqrt(c))) = ∫(1/sqrt(c)) dv = v + C2 = (x/sqrt(c)) + C2
where C2 is the constant of integration.
Substituting these integrals back into the original equation, we get:
(1/2) ln|y^2 - c| = +(x/sqrt(c)) + C
where C = C1 + C2 is the overall constant of integration.
Taking the exponential of both sides, we get:
|y^2 - c|^(1/2) = e^(2x/sqrt(c) + C)
Since the sign of the square root depends on the sign of the quantity under the absolute value, we can write the solution in two cases:
1. y^2 - c > 0:
Squaring both sides, we get:
y^2 - c = e^(4x/sqrt(c) + 2C)
Solving for y, we get:
y = +(e^(4x/sqrt(c) + 2C) + c)^(1/2)
2. y^2 - c < 0:
Squaring both sides and multiplying by -1, we get:
y^2 - c = -e^(4x/sqrt(c) + 2C)
Solving for y, we get:
y = +(c - e^(4x/sqrt(c) + 2C))^(1/2)
Therefore, the general solution to the differential equation (dy/dx)/(((y^2) - c)^(1/2)) = +/-(1/(sqrt(c))) is given by:
y =
{
+(e^(4x/sqrt(c) + 2C) + c)^(1/2) for y^2 - c > 0
+(c - e^(4x/sqrt(c) + 2C))^(1/2) for y^2 - c < 0
}
where C is the constant of integration.