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Solve the differential equation: e^(2y) * (dy/dx) + tan(x) = 0, given that y = 0 when x = 0. Give your answer in the form y = f(x).

This is a question taken from a core 4 paper and is a typical example of a differential equation question.

The first thing to notice about this equation is that it is "separable", meaning we can rearrange it to get 

e^(2y) dy = - tan(x) dx

Now we can solve this by integrating both sides. We know how to integrate the left hand side, and we get (1/2)e^(2y), but how can we integrate -tan(x)?

To see how we can do this, we write

-tan(x) = -sin(x) / cos(x)

Then, we realise that the numerator is the derivative of the denominator, and so integrating -tan(x) gives ln(|cos(x)|) + C, where C is the constant of integration. 

So, we now have that

(1/2)e^(2y) = ln(|cos(x)|) + C

Now we apply the condition that y(x=0) = 0, giving

1/2 = C

Subbing this in, we have

(1/2)e^(2y) = ln(|cos(x)|) + 1/2

Therefore 

e^(2y) = 2ln(|cos(x)|) + 1

The question asked for the answer to be written in the form y = f(x), and so we need to get the y out of the exponent, which we can do by taking ln of both sides to give

2y = ln( 1 + 2ln(|cos(x)|)  )

And so the final answer is

y = (1/2) ln( 1 + 2ln(|cos(x)|) )

Dominic D. GCSE Further Mathematics  tutor, A Level Further Mathemati...

3 months ago

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