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.
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.
The slope of the characteristic lines is 2/5. This is derived from the direction vector (5,2), where the change in y is 2 and the change in x is 5, so dy/dx=2/5.
Conditions: The direction vector is (5,2).; The characteristic lines are straight lines in the xy-plane.
The slope of the characteristic lines is 2/5. This is derived from the direction vector (5,2), where the change in y is 2 and the change in x is 5, so dy/dx=2/5.
Conditions: The direction vector is (5,2).; The characteristic lines are straight lines in the xy-plane.
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.
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.