Question

${\displaystyle (2ax+x^2)\frac{dy}{dx}=a^2+2ax}$

Answer 1

$$(2ax+x^2)\frac{dy}{dx}=a^2+2ax$$

$$\frac{dy}{dx}=\frac{a^2+2ax}{2ax+x^2}=\frac{a(a+2x)}{x(2a+x)}$$

$$\text{Let }x=2a{\tan }^2\theta \Rightarrow dx=4a\tan \theta {\sec }^2\theta d\theta$$

$$\frac{dy}{dx}=\frac{a(a+4ata{n}^2\theta )}{2a{\tan }^2\theta \left(2a\right)(1+{\tan }^2\theta )}$$

$$\int dy=\int \frac{a(1+4{\tan }^2\theta )}{2{\tan }^2\theta \left(2a\right)({\sec }^2\theta )}dx$$

$$\int dy=\int \frac{a(1+4{\tan }^2\theta )}{2{\tan }^2\theta \left(2a\right)({\sec }^2\theta )}\left(4a\right)\tan \theta {\sec }^2\theta d\theta$$

$$=\int \frac{a(1+4{\tan }^2\theta )}{\tan \theta }d\theta$$

$$=a\int (\frac{1}{\tan \theta }+4\tan \theta )d\theta$$

$$y=a\int \cot \theta +4tan\theta d\theta$$

$$y=a[\log \sin \theta +4(-\log \cos \theta )]+c$$

$$y=a[\log \sin \theta -4\log \cos \theta ]+c$$

$$\text{As, }x=2a{\tan }^2\theta \Rightarrow \tan \theta =\sqrt{\frac{x}{2a}}$$

$$y=a\log \left(\frac{\sin \theta }{{\cos }^4\theta }\right)+c$$

$$=a\log \left(\frac{\tan \theta }{{\cos }^3\theta }\right)+c$$

$$=a\log (\sqrt{\frac{x}{2a}}\times {\left(\sqrt{\frac{x+2a}{2a}}\right)}^3)+c$$

$$y=a\log \left(\frac{x^{\frac{1}{2}}{(x+2a)}^{\frac{3}{2}}}{4a^2}\right)+c$$

$$y+C=\frac{a}{2}(\log x+3\log (x+2a))\text{ where }C=c-a\log \left(4a^2\right)$$