Tricky Differential Equation Problem
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, 2020Comments

17andLearning about 3 years
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)