What is the slope of the characteristic lines for the direction vector (5,2)?

The equation (5,2)·∇$u=0$ implies that the directional derivative of u in the direction of the vector (5,2) is zero. This means that the rate of change of u is zero when moving in that specific direction.

Conditions: The direction vector is (5,2).; ∇u is the gradient of u.

Rewriting the PDE as a dot product with the gradient allows the equation to be interpreted as a statement about directional derivatives. Specifically, it shows that the directional derivative of u in the direction (5,2) is zero, which implies that u is constant along curves in that direction.

Conditions: The PDE is $5u_x+2u_y=0$.; ∇u is the gradient vector ($u_x$,$u_y$).

The general solution is $u(x,y) = f(y - (2/5)x)$, where f is an arbitrary differentiable function of one variable. This form arises because u is constant along the characteristic curves $y = (2/5)x + c$, so u depends only on the invariant $c = y - (2/5)x$.

Conditions: u is constant along the characteristic curves.; f is an arbitrary function of one variable.

u is constant along curves in the direction (5,2) because the directional derivative of u in that direction is zero. If the rate of change of a function is zero in a specific direction, the function's value does not change as one moves along any curve tangent to that direction.

Conditions: (5,2)·∇$u=0$.; The curves considered are tangent to the direction (5,2).

Integrate the ordinary differential equation $dy/dx = 2/5$ with respect to x. This yields $y = (2/5)x + c$, where c is an arbitrary constant representing the y-intercept.

Conditions: The ODE is $dy/dx = 2/5$.; c is an arbitrary real constant.