We #160;have,dydx=x2y+xThis #160;is #160;a #160;homogeneous #160;differential #160;equation.Putting #160;y=vx #160;and #160;dydx=v+xdvdx, #1 ...

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Integration of Irrational Algebraic Fractions - 1

dydx =x 2 y+x

Question

$\frac{d y}{d x} = \frac{x}{2 y + x}$

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Solution

$We have, \frac{d y}{d x} = \frac{x}{2 y + x} This is a homogeneous differential equation . Putting y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x}, we get v + x \frac{d v}{d x} = \frac{x}{2 v x + x} \Rightarrow v + x \frac{d v}{d x} = \frac{1}{2 v + 1} \Rightarrow x \frac{d v}{d x} = \frac{1}{2 v + 1} - v \Rightarrow x \frac{d v}{d x} = \frac{1 - 2 v^{2} - v}{2 v + 1} \Rightarrow \frac{2 v + 1}{1 - 2 v^{2} - v} d v = \frac{1}{x} d x Integrating both sides, we get \int \frac{2 v + 1}{1 - 2 v^{2} - v} d v = \int \frac{1}{x} d x \Rightarrow \int \frac{2 v + 1}{2 v^{2} + v - 1} d v = - \int \frac{1}{x} d x \Rightarrow \int \frac{2 v + 1}{2 v (v + 1) - 1 (v + 1)} d v = - \int \frac{1}{x} d x \Rightarrow \int \frac{2 v + 1}{(2 v - 1) (v + 1)} d v = - \int \frac{1}{x} d x . . . . . (1) Solving left hand side integral of (1), we get Using partial fraction, Let \frac{2 v + 1}{(2 v - 1) (v + 1)} = \frac{A}{(2 v - 1)} + \frac{B}{(v + 1)} ∴ A + 2 B = 2 . . . . . (2) And A - B = 1 . . . . . (3) Solving (2) and (3), we get A = \frac{4}{3} and B = \frac{1}{3} ∴ \int \frac{2 v + 1}{(2 v - 1) (v + 1)} d v = \frac{4}{3} \int \frac{1}{2 v - 1} d v + \frac{1}{3} \int \frac{1}{v + 1} d v = \frac{4}{3 \times 2} \log |2 v - 1| + \frac{1}{3} \log |v + 1| + \log C From (1), we get \frac{2}{3} \log |2 v - 1| + \frac{1}{3} |v + 1| + \log C = - \log |x| + \log C_{1} \Rightarrow \log \{|{(2 v - 1)}^{2}| |v + 1|\} = - 3 \log |x| + \log C_{2} \Rightarrow \log \{|{(2 v - 1)}^{2}| |v + 1|\} = \log |\frac{{C_{2}}^{3}}{x^{3}}| \Rightarrow {(2 v - 1)}^{2} (v + 1) = \frac{{C_{2}}^{3}}{x^{3}} Putting v = \frac{y}{x}, we get \Rightarrow {(\frac{2 y - x}{x})}^{2} (\frac{y + x}{x}) = \frac{{C_{2}}^{3}}{x^{3}} \Rightarrow (x + y) {(2 y - x)}^{2} = k$

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