Answer to Question #275541 in Calculus for dandelion

Question

Apply separation of variable to solve

x²u_xy+9y²u=0

Answer 1

$x^2u_{xy}+9y^2u=0\ ...(i) \\ Let\ u=f(x)g(y) \\u_x=f'(x)g(y) \\u_y=f(x)g'(y)$

Substituting in (i):

$x^2f'g'+9y^2fg=0 \\ x^2f'g'=-9y^2fg \\ \dfrac{x^2f'}{f}=\dfrac{-9y^2g}{g'}=\lambda\ (\lambda>0)$

Consider $\dfrac{x^2f'}{f}=\lambda\ \& \ \dfrac{-9y^2g}{g'}=\lambda$

$\Rightarrow\dfrac{df}{f}=\dfrac{\lambda dx}{x^2}\ \&\ \dfrac{g'}{g}=\dfrac{-9y^2}{\lambda} \\ \Rightarrow\log f=\dfrac{-\lambda}{x}+C_1\ \&\ \dfrac{dg}{g}=\dfrac{-9y^2dy}{\lambda} \\\Rightarrow f=e^{\frac{-\lambda}{x}+C_1}\ \&\ \log g=\dfrac{-9}{\lambda}.\dfrac{y^3}{3}+C_2 \\\Rightarrow f=Ae^{\frac{-\lambda}{x}}\ \&\ \log g=\dfrac{-3y^3}{\lambda}+C_2$

$\Rightarrow g=e^{\frac{-3y^3}{\lambda}+C_2} \\ \Rightarrow g=Be^{\frac{-3y^3}{\lambda}} \\\therefore u(x,y)=Ae^{\frac{-\lambda}{x}}.Be^{\frac{-3y^3}{\lambda}} \\\Rightarrow u(x,y)=Ce^{\frac{-\lambda}{x}-\frac{3y^3}{\lambda}} \ \ [C=A.B]$