A lvl H2 Maths: Differential Equations

Question from http://www.sgforums.com/forums/2297/topics/404564

Find the particular solution for
(4x+xy²)dx +(y+yx²)dy=0 , given that y(1)=2.

*A poly maths question, but nevertheless good for A levels as well!

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Answer:
(4x+xy²) dx = -(y+yx²) dy
$\frac{dy}{dx}=-\frac{4x+xy^{2}}{y+yx^{2}}$
$\frac{dy}{dx}=-\frac{x(4+y^{2})}{y(1+x^{2})}$
$\frac{y}{4+y^{2}}dy=-\frac{x}{1+x^{2}}dx$
$\int \frac{y}{4+y^{2}}dy=-\int \frac{x}{1+x^{2}}dx$
$\frac{1}{2}\int \frac{2y}{4+y^{2}}dy=-\frac{1}{2}\int \frac{2x}{1+x^{2}}dx$
$\frac{1}{2}\ln (4+y^{2})=-\frac{1}{2}\ln (1+x^{2})+c$
ln (4 + y²) = -ln (1 + x²) + C, where C = 2c
ln (4 + y²) + ln (1 + x²) = C
(4 + y²) (1 + x²) = A, where A = e^c

When x = 1, y = 2
Hence,
A = (4 + 4) * (1 + 1) = 16

Thus,
(4 + y²) (1 + x²) =16

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A lvl H2 Maths: Differential Equations