Show that every solution curve for the equation approaches the line as .

Before showing what is asked for above I would like to show the derivation of a formula that will come in handy for solving this problem. The derived formula is used for finding the solution to a linear first-order equation.

I begin by writing down a linear first-order equation in the form on an interval on which the coefficient functions and are continuous. I next multiply each side of the equation by the integrating factor . An integrating factor for a differential equation is a function, such that the multiplication of each side of the differential equation by yields an equation in which each side is recognizable as a derivative. After the multiplication I obtain the equation . Now, because , the left-hand side is the derivative product so the equation becomes . Integration of both sides of the equation gives me . Finally, I can solve for which will give me the general solution for linear first-order equations which will be useful for the initial proof. The general solution is .

Now, for the proof I start by finding the integrating factor. After writing the given equation in the form I find that is which implies that which equals . Now, I can multiply each side of the equation by which gives me . Because of the product rule of derivates this equation is equivalent to . After integrating each side of the equation, I obtain . For the right side of the equation, I am going to have to apply the Integration by Parts technique which is of the formula . To use this formula, I must choose the right values for and . I choose , and which implies that and . Now, I can substitute the appropriate variables into the Integration by Parts formula to obtain the equation . I now have a final integration to do on the right side of the equation and after doing so I obtain . I can then solve for to obtain . Finally, I can take , and in this limit the term goes to because the exponential gets larger and larger as gets larger and larger which causes the fraction to essentially approach . This means that approaches the line which is what was asked to prove.

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