- #1
lackrange
- 19
- 0
Problem: Find the characteristics of
xyu_x+(2y^2-x^6)u_y=0
So I rewrote this as u_x+\frac{2y^2-x^6}{xy}u_y=0 and then set this as <br /> \frac{du}{dx}=0\implies \frac{dy}{dx}=\frac{2y^2-x^6}{xy}
I solved this, and found that the characteristics were \frac{y^2+x^6}{x^4}=C
where C is a constant, and u is constant along this curve. Now the problem says consider the initial condition u(x,α x^n)=x^2,\;\;n\in \mathbb{N}\;\;α>0,
for what α>0 does the problem have a solution? For what α > 0 is the solution uniquely? Your answer may depend on n (Try n=1, n=2 etc.).
So I wrote αx^n=\frac{y^2+x^6}{x^4} and solved for α, but I don't think this is what I am suppose to do, can someone help me please?
xyu_x+(2y^2-x^6)u_y=0
So I rewrote this as u_x+\frac{2y^2-x^6}{xy}u_y=0 and then set this as <br /> \frac{du}{dx}=0\implies \frac{dy}{dx}=\frac{2y^2-x^6}{xy}
I solved this, and found that the characteristics were \frac{y^2+x^6}{x^4}=C
where C is a constant, and u is constant along this curve. Now the problem says consider the initial condition u(x,α x^n)=x^2,\;\;n\in \mathbb{N}\;\;α>0,
for what α>0 does the problem have a solution? For what α > 0 is the solution uniquely? Your answer may depend on n (Try n=1, n=2 etc.).
So I wrote αx^n=\frac{y^2+x^6}{x^4} and solved for α, but I don't think this is what I am suppose to do, can someone help me please?