#### 2.1.75 $$x u_t - t u_x = 0$$

problem number 75

From Math 5587 midterm I, Fall 2016, practice exam, problem 6.

Solve for $$u(x,t)$$ with $$x>0,t>0$$ and $$u(x,0)=x^2$$ \begin {align*} x u_t - t u_x = 0 \end {align*}

Mathematica

$\left \{\left \{u(x,t)\to c_1\left (\frac {1}{2} \left (t^2+x^2\right )\right )\right \}\right \}$

Maple

$u \left (x , t\right ) = \mathit {\_F1} \left (t^{2}+x^{2}\right )$

Hand solution

Solve $$xu_{t}-tu_{x}=0$$. Using the method of characteristics$\frac {dt}{x}=-\frac {dx}{t}=\frac {du}{0}$ From the ﬁrst pair of equations we obtain $$tdt=-xdx$$ or $$\frac {t^{2}}{2}=-\frac {x^{2}}{2}+C_{1}$$ or $$C_{1}=\frac {t^{2}}{2}+\frac {x^{2}}{2}$$. From $$du=0$$ then $$u=C_{2}$$. But $$C_{2}=F\left ( C_{1}\right )$$ where $$F$$ is arbitrary function. Hence $u=F\left ( \frac {t^{2}}{2}+\frac {x^{2}}{2}\right )$

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