Question

Find the general solution to the given partial differential equation and use it to find the solution satisfying the given initial data.

$\frac{\partial u}{\partial x}=-(2x+y)\frac{\partial u}{\partial y}$

$u(0,y)=1+y^2$

Answer 1

$\frac{dx}{1}=\frac{dy}{2x+y}$

$(2x+y)dx-dy=0$

$\mu(x)=e^{-x}$

$e^{-x}y'-e^{-x}y=2xe^{-x}$

$\frac{d}{dx}(ye^{-x})=2xe^{-x}$

$ye^{-x}=\int 2xe^{-x}dx=-2e^{-x}(x+1)+c$

$u(x,y)=c$

$ye^{-x}+2e^{-x}(x+1)=c$

$u(x,y)=ye^{-x}+2e^{-x}(x+1)$

$u(0,y)=y+2=y^2+1$

$y^2-y-1=0$

$y_1=\frac{1-\sqrt{5}}{2}$

$u_1(x,y)=\frac{1-\sqrt{5}}{2}e^{-x}+2e^{-x}(x+1)$

$y_2=\frac{1+\sqrt{5}}{2}$

$u_2(x,y)=\frac{1+\sqrt{5}}{2}e^{-x}+2e^{-x}(x+1)$