# Tricky Differential Equation Problem

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## Solution 1

Rewrite the equation as: $$\frac{dx}{dy} + x = e^{-y} \sec^2y$$

This can now be tackled with an 'Integrating Factor' which in this case is $e^y$.

So, multiply both sides by $e^y$: $$e^y\frac{dx}{dy} + xe^y = \sec^2y$$

The LHS is nothing but the derivative of $xe^y$ with respect to $y.$

We have $$\frac{d(xe^y)}{dy}= \sec^2y$$

Integrate both sides with respect to $y$: $$xe^y=\tan(y)+c$$

Thus, $$x=e^{-y}\tan(y)+ce^{-y}$$

where $c$ is an arbitrary constant.

Note: Any equation of the form $$\frac{dx}{dy} + xP(y) = Q(y)$$ can be tackled using an Integrating Factor.

## Solution 2

Hint: Rewrite/Simplify it as:

$$e^y~dx + (x e^y - \sec^2 y)~ dy = 0$$

Now, check this as being an Exact Equation to proceed.

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### 17andLearning

Updated on October 08, 2020

I am unsure of how to tackle the following differential equation:

$$dx+ x\,dy = e^{-y}\sec^2y\,dy$$

I have done the following so far:

$$dx + x\,dy = e^{-y} \sec^2y \, dy$$

$$=>dx = e^{-y} \sec^2y \, dy - x \, dy$$

$$=>dx = (e^{-y} \sec^2y - x) \, dy$$

$$=>dx/dy = (e^{-y}) \sec^2y - x$$

Is this the correct approach? How can the problem be solved after this?

(Not homework: preparing for a test)