MHB Differential Equation Challenge

[Dethrone]

Find $y(x)$ to satisfy $$ y(x)=y'(x)+\int e^{2x}y(x) \, dx+\lim_{{x}\to{-\infty}}y(x)$$ given $$\lim_{{x}\to{0}}y(x)=0$$ and $$\lim_{{x}\to{\ln\left({\pi/2}\right)}}y(x)=1.$$

Spoiler

Source: Calc I exam

 

[topsquark]

Spoiler

Calc I?? How many Red Bulls did your professor drink while he was writing the exam?

-Dan

 

[Siron]

I tried something:

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Here's an attempt.
$$y(x) = y'(x) + \int e^{2x} y(x)dx + \lim_{x \to - \infty} y(x)$$
Writing $\int y'(x)dx = y(x)$ the above becomes
$$\int [y'(x)-e^{2x} y(x)]dx = y'(x) + \lim_{x \to -\infty} y(x)$$
Assuming the second derivative of $y(x)$ exists and assuming that $\lim_{x \to - \infty} y(x) < \infty$ (otherwise the equality would nog make sense) differentiating both sides leads to
$$y'(x)-e^{2x}y(x) = y''(x)$$
The above differential equation is not that hard to solve, the solution is given by
$$y(x) = c_1\sin(e^x)+c_2\cos(e^x)$$
where $c_1,c_2 \in \mathbb{R}$ need to be determined. We can do this by the given conditions. More precisely,

$$\lim_{x \to \ln\left(\frac{\pi}{2}\right)} y(x) = c_1 = 1$$
the second condition gives
$$\lim_{x \to 0} y(x) = \sin(1)+c_2 \cos(1)$$
therefore the solution is

$$y(x) = \sin(e^x) - \frac{\sin(1)}{\cos(1)}\cos(e^x)$$

 

[Dethrone]

Sorry Siron for forgetting to get back to you. That is the correct answer. :D Thanks for participating.