Solution of 1st order linear ODE

Question

For $\alpha \in \mathbb{R}$, let $y_{\alpha}(x)$ be the solution of the differential equation $$\frac{dy}{dx} + 2 y = \frac{1}{1+x^2},\text{satisfying} \space y(0)=\alpha.$$Then prove that $\lim\limits_{x\to\infty}y_{\alpha}(x)=0$.

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Since the above differential equation is a first order linear differential equation with integrating factor $e^{2x}$. Therefore its solution is given by, $$y_{\alpha}(x) = e^{-2x} \left(\int \left(\frac{e^{2x}}{1+x^2}\right) \rm d x + c\right)$$ Now I am stuck with this integral. I have tried integration by parts but because of exponential and $\frac{1}{1+x^2}$ term it is not useful.I tried substitutions but did not get any help. Kindy someone help me to solve this integral or if there is any other method to solve it then please share.

Answer 1

You don't need to do the integral. Put the $e^{-2x}$ in the denominator and use L'hospital's Rule and Fundamental Theorem of Calculus.

$$\lim_{x \to \infty} \frac{\int_0^x \frac{e^{2t}}{1+t^2}\; dt}{e^{2x}} =\lim_{x\to \infty} \frac{ \frac{e^{2x}}{1+x^2}}{2e^{2x}} =\lim_{x\to \infty} \frac{1}{2(1+x^2)} = 0. $$

Answer 2

By definition of the exponential integral function

$$\int \frac{e^{-x}}{x} d \ x = Ei[x]+c.$$

By the series it's a $\log(x)$ plus an entier function $$\mathop{Ei}(x) \ = \ \log x + \sum_{n=1}^\infty \frac{(-x)^{n}}{n \ n!}$$

For your problem use partial fraction decomposition

$$\frac{1}{1+x^2} = \frac{1}{2}\left( \frac{1}{x +i } + \frac{1}{x-i} \right)$$

and set the integration path and boundaries such that for $x=0$ the value a is assumed.

Answer 3

Here is another proof: \begin{align} y_{\alpha}(x)&=e^{-2x}\left(\alpha+\int_0^x\frac{e^{2t}}{1+t^2}\,dt\right) \\ &=e^{-2x}\left(\alpha+\int_0^{x/2}\frac{e^{2t}}{1+t^2}\,dt +\int_{x/2}^{x}\frac{e^{2t}}{1+t^2}\,dt\right), \tag{1} \end{align} hence $$ |y_{\alpha}(x)|\leq e^{-2x}\left(|\alpha|+\left|\int_0^{x/2}\frac{e^{2t}}{1+t^2}\,dt\right|+\left|\int_{x/2}^{x}\frac{e^{2t}}{1+t^2}\,dt\right|\right). \tag{2} $$ For $x>0$ we have $$ \left|\int_0^{x/2}\frac{e^{2t}}{1+t^2}\,dt\right|<\int_0^{x/2}e^{2t}\,dt<\frac{e^x}{2} \tag{3} $$ and $$ \left|\int_{x/2}^x\frac{e^{2t}}{1+t^2}\,dt\right|<\frac{1}{1+(x/2)^2}\int_{x/2}^{x}e^{2t}\,dt<\frac{2e^{2x}}{4+x^2}. \tag{4} $$ Plugging $(3)$ and $(4)$ into $(2)$ we get $$ |y_{\alpha}(x)|<e^{-2x}|\alpha|+\frac{e^{-x}}{2}+\frac{2}{4+x^2}\qquad(x>0), \tag{5} $$ from which follows $$ \lim_{x\to+\infty}y_{\alpha}(x)=0.\qquad\square \tag{6} $$