Differential Equations Proof

Show that every solution curve for the equation y'=x-y approaches the line y=x-1 as x\to +\infty .

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 \frac{dy}{dx} + P(x)y=Q(x) on an interval on which the coefficient functionsP(x) and Q(x) are continuous. I next multiply each side of the equation by the integrating factor \rho(x)=e^{\int P(x) dx} . An integrating factor for a differential equation is a function, \rho(x,y) such that the multiplication of each side of the differential equation by \rho(x,y) yields an equation in which each side is recognizable as a derivative. After the multiplication I obtain the equation e^{\int P(x) dx}\frac{dy}{dx}+P(x)e^{\int P(x) dx}y=Q(x)e^{\int P(x) dx} . Now, because D_{x}[\int P(x) dx]=P(x) , the left-hand side is the derivative product y(x)e^{\int P(x) dx} so the equation becomes D_{x}[y(x)e^{\int P(x) dx}]=Q(x)e^{\int P(x) dx } . Integration of both sides of the equation gives me y(x)e^{\int P(x) dx}=\int(Q(x)e^{\int P(x) dx})dx + C . Finally, I can solve for y(x) which will give me the general solution for linear first-order equations which will be useful for the initial proof. The general solution is y(x)=e^{-\int P(x)dx}[\int (Q(x)e^{\int P(x) dx}) dx +C] .

Now, for the proof I start by finding the integrating factor. After writing the given equation in the form \frac{dy}{dx} + P(x)y=Q(x) I find that P(x) is 1 which implies that \rho =e^{\int 1 dx} which equals e^{x} . Now, I can multiply each side of the equation by \rho which gives me e^{x}\frac{dy}{dx} +ye^{x}=xe^{x} . Because of the product rule of derivates this equation is equivalent to D_{x}[e^{x}y]=xe^{x} . After integrating each side of the equation, I obtain e^{x}y = \int xe^{x} dx . For the right side of the equation, I am going to have to apply the Integration by Parts technique which is of the formula uv - \int v du . To use this formula, I must choose the right values for u and v . I choose u =x , and v = e^{x} which implies that du = 1 dx and dv = e^{x} . Now, I can substitute the appropriate variables into the Integration by Parts formula to obtain the equation e^{x}y = xe^{x} - \int e^{x} dx . I now have a final integration to do on the right side of the equation and after doing so I obtain e^{x}y = xe^{x} - e^{x} + C . I can then solve for y(x) to obtain y = x - 1 + \frac{C}{e^{x}} . Finally, I can take \displaystyle \lim_{x \to + \infty} , and in this limit the term \frac{C}{e^{x}} goes to 0 because the exponential gets larger and larger as x gets larger and larger which causes the fraction to essentially approach 0 . This means that y(x) approaches the line x - 1 which is what was asked to prove.

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