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Property 4

2 a x+x2 d y ...

Question

(2ax + x²) $\frac{d y}{d x}$ = a² + 2ax

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Solution

$\begin{array}{l} (2 a x + x^{2}) \frac{d y}{d x} = a^{2} + 2 a x \\ \frac{d y}{d x} = \frac{a^{2} + 2 a x}{2 a x + x^{2}} = \frac{a (a + 2 x)}{x (2 a + x)} \\ Let x = 2 a \tan^{2} θ \Rightarrow d x = 4 a \tan θ \sec^{2} θ d θ \\ \frac{d y}{d x} = \frac{a (a + 4 a \tan^{2} θ)}{2 a \tan^{2} θ (2 a) (1 + \tan^{2} θ)} \\ \int d y = \int \frac{a (1 + 4 \tan^{2} θ)}{2 \tan^{2} θ (2 a) (\sec^{2} θ)} d x \\ \int d y = \int \frac{a (1 + 4 \tan^{2} θ)}{2 \tan^{2} θ (2 a) (\sec^{2} θ)} (4 a) \tan θ \sec^{2} θ d θ \\ = \int \frac{a (1 + 4 \tan^{2} θ)}{\tan θ} d θ \\ = a \int (\frac{1}{\tan θ} + 4 \tan θ) d θ \\ y = a \int \cot θ + 4 t a n θ d θ \\ y = a [\log \sin θ + 4 (- \log \cos θ)] + c \\ y = a [\log \sin θ - 4 \log \cos θ] + c \\ As, x = 2 a \tan^{2} θ \Rightarrow \tan θ = \sqrt{\frac{x}{2 a}} \\ y = a \log (\frac{\sin θ}{\cos^{4} θ}) + c \\ = a \log (\frac{\tan θ}{\cos^{3} θ}) + c \\ = a \log (\sqrt{\frac{x}{2 a}} \times {(\sqrt{\frac{x + 2 a}{2 a}})}^{3}) + c \\ y = a \log (\frac{x^{\frac{1}{2}} {(x + 2 a)}^{\frac{3}{2}}}{4 a^{2}}) + c \end{array} y + C = \frac{a}{2} (\log x + 3 \log (x + 2 a)) where C = c - a \log (4 a^{2})$

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\frac{a + \sqrt{a^{2} - 2 a x}}{\sqrt{a^{2} - 2 a x}} = b

Q. Find the solution of

(2 a x + x^{2}) \frac{d y}{d x} = a^{2} + 2 a x

a

b

are rational numbers and

b

is not a perfect square, then the quadratic equation with rational coefficients whose one root is

\frac{1}{a + \sqrt{b}}

(a^{2} - b^{2}) x^{2} + 2 a x + 1

x^{2} - 2 a x + (a^{2} - b^{2}) = 0

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