Question

Solve the following initial value problem:
x (x² + 3y²) dx + y (y² + 3x²) dy = 0, y (1) = 1

Answer 1

\[x( x^2 + 3 y^2 )dx + y( y^2 + 3 x^2 )dy = 0, y(1) = 1\]
\[ \frac{dy}{dx} = \frac{- x( x^2 + 3 y^2 )}{y( y^2 + 3 x^2 )}\]
it is a homogeneous equation . Put y = vx
\[\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ So, }v + x\frac{dv}{dx} = - \frac{x( x^2 + 3 v^2 x^2 )}{vx( v^2 x^2 + 3 x^2 )}\]
\[x\frac{dv}{dx} = - \frac{(1 + 3 v^2 )}{v( v^2 + 3)} - 3\]
\[ = \frac{- 1 - 3 v^2 - v^4 - 3 v^2}{v( v^2 + 3)}\]
\[x\frac{dv}{dx} = \frac{- v^4 - 6 v^2 - 1}{v( v^2 + 3)}\]
\[\frac{v( v^2 + 3)}{v^4 + 6 v^2 + 1}dv = - \frac{dx}{x}\]
\[\int\frac{4 v^3 + 12v}{v^4 + 6 v^2 + 1}dv = - 4\int\frac{dx}{x}\]
\[\log\left| v^4 + 6 v^2 + 1 \right| = \log\left| \frac{c}{x^4} \right|\]
\[\left| v^4 + 6 v^2 + 1 \right| = \left| \frac{c}{x^4} \right|\]
\[\left| y^4 + 6 y^2 x^2 + x^4 \right| = \left| c \right| . . . . (1)\]
\[\text{ put }y = 1, x = 1\]
\[(1 + 6 + 1) = c \Rightarrow c = 8\]
\[\text{ put }c = 8\text{ in equation }(1), \]
\[( y^4 + x^4 + 6 x^2 y^2 ) = 8\]