Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 26

$$y(x)=\frac{\tan ^{-1} x-\pi/4}{\sqrt{1+x^{2}}}$$

Work Step by Step

Given$$y^{\prime}+\frac{x}{1+x^{2}} y=\frac{1}{\left(1+x^{2}\right)^{3 / 2}}, \quad y(1)=0$$ This is a linear equation with $p(x) =\dfrac{x}{1+x^{2}}\ \ q(x) =\dfrac{1}{\left(1+x^{2}\right)^{3 / 2}}$, so \begin{align*} \mu(x)&=e^{\int p(x)dx}\\ &=e^{\int \frac{x}{1+x^{2}} d x}\\ &=e^{\frac{1}{2}\ln (1+ x^2)}\\ &= (1+ x^2)^{1/2} \end{align*} Then \begin{align*} y\mu(x) &=\int \mu(x)q(x)dx\\ (1+ x^2)^{1/2} y &=\int (1+ x^2)^{1/2} \dfrac{1}{\left(1+x^{2}\right)^{3 / 2}}dx\\ &= \int \dfrac{1}{1+x^{2}}dx\\ &=\tan^{-1}x+C \end{align*} Then $$y(x)=\frac{\tan ^{-1} x+C}{\sqrt{1+x^{2}}}$$ Since $y(1)=0$, then $C= -\pi /4$, and hence $$y(x)=\frac{\tan ^{-1} x-\pi/4}{\sqrt{1+x^{2}}}$$

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