How do you find the solution to the differential equation #dy/dx=cos(x)/y^2# where #y(π/2)=0#?
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To find the solution to the differential equation (\frac{dy}{dx} = \frac{\cos(x)}{y^2}) with the initial condition (y(\frac{\pi}{2}) = 0), follow these steps:

Separate the variables by multiplying both sides by (y^2) and dividing both sides by (\cos(x)), yielding (y^2 dy = \frac{1}{\cos(x)} dx).

Integrate both sides with respect to their respective variables. This gives (\int y^2 dy = \int \sec(x) dx).

Integrate (\int \sec(x) dx) to get (\ln\sec(x) + \tan(x) + C_1), where (C_1) is the constant of integration.

Integrate (\int y^2 dy) to get (\frac{1}{3}y^3 + C_2), where (C_2) is the constant of integration.

Combine the constants of integration into one constant, let's call it (C), so the solution becomes (\frac{1}{3}y^3 + \ln\sec(x) + \tan(x) = C).

Apply the initial condition (y(\frac{\pi}{2}) = 0) to find the value of the constant (C).

Substitute (x = \frac{\pi}{2}) and (y = 0) into the solution. This gives (0 + \ln\sec(\frac{\pi}{2}) + \tan(\frac{\pi}{2}) = C), which simplifies to (0 + \ln1 + \infty = C).

Since (\tan(\frac{\pi}{2}) = \infty), the absolute value of (\sec(\frac{\pi}{2}) + \tan(\frac{\pi}{2})) is (\infty), so (\ln1 + \infty = \infty).

Therefore, the constant (C) is (+\infty).

The solution to the differential equation with the initial condition is (\frac{1}{3}y^3 + \ln\sec(x) + \tan(x) = +\infty).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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