Solve $xu_x+(x+2y-2)u_y=x+y, \space x>0, \quad u(1,y)=y$
My attempt:
- $x'(t)=x(t) \space \Rightarrow \space x(t)=c_1 e^t$
- $y'(t)=x(t)+2y(t)-2 \space \Rightarrow \space y'(t)-2y(t)=c_1 e^t-2 \space \Rightarrow \space y(t)=-c_1e^t +c_2e^{2t}+1$
- $u'(t)=x(t)+y(t) \space \Rightarrow \space u'(t)=c_2e^{2t}+1 \space \Rightarrow \space u(t)=\frac {1}{2} c_2 e^{2t} +t+c_3$
Now using the initial condition:
- $x(0)=c_1=1$
- $y(0)=-1+c_2+1=c_2=r$
- $u(0)=\frac{r}{2} +c_3=r \Rightarrow c_3=\frac{r}{2}$
Substituting back in the equations:
- $x(t)=c_1 e^t= e^t$
- $y(t)=-c_1e^t +c_2e^{2t}+1=-e^t +re^{2t}+1$
- $u(t)=\frac{r}{2}e^{2t}+t+\frac{r}{2}$
Now:
$t=ln(x) \space \Rightarrow \space y=-e^{ln(x)} +re^{2ln(x)}+1 \space \Rightarrow \space y=-x+r x^2 +1 \space \Rightarrow \space r=\frac{y+x-1}{x^2}$
substituting in $u(t)$
$u(t)=\frac{r}{2}e^{2t}+t+\frac{r}{2} \space \Rightarrow \space u(x,y)=\frac{y+x-1}{2}+ln(x)+\frac{y+x-1}{2 x^2}$
This is unfortunately wrong, can someone help me find where the mistake is?