Question

Solve the following IVP, and check the non-characteristic condition first: UU_x + yU_y = x U = 2s on a characteristic curve gamma: x = s, y = s, s element of R.

Answer 1

Answer on question 38939 – Math – Differential Calculus

Solve the following IVP, and check the non-characteristic condition first: $UU_x + yU_y = xU = 2s$ on a characteristic curve gamma: $x = s, y = s, s$ element of R.

Solution

$$\left\{ \begin{array}{c} u u _ {x} + y u _ {y} = x \\ u = 2 s \\ x = y = s \end{array} \right.$$

First independent integrals are

$$\frac {d x}{u} = \frac {d y}{y} = \frac {d u}{x}$$

Therefrom we get

$$\frac {d x}{u} = \frac {d u}{x} \Rightarrow \int x d x = \int u d u \Rightarrow u ^ {2} = x ^ {2} + c _ {1} \Rightarrow c _ {1} = u - x; \quad (*)$$ $$\frac {d y}{y} = \frac {d u}{x} \Rightarrow \frac {u}{x} = \ln | y | + c _ {2} \Rightarrow c _ {2} = \frac {u}{x} - \ln | y |.$$

Considering the initial conditions we get

$$c _ {1} = 2 s - s = s, \quad c _ {2} = \frac {2 s}{s} - \ln | s | = 2 - \ln | s |$$

Or we can write

$$c _ {2} = 2 - \ln | c _ {1} |$$

Substitute (*) and (**) into this equation

$$\frac {u}{x} - \ln | y | = 2 - \ln | u - x |$$ $$\frac {u}{x} = 2 + \ln | \frac {y}{u - x} |$$

Answer: $\frac{u}{x} - 2 - \ln \left| \frac{y}{u - x} \right| = 0.$