Toronto Math Forum
APM346-2018S => APM346--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 06:56:34 PM
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Consider the first order equation:
\begin{equation}
u_t + 3(t^2-1) u_x = 6t^2.
\tag{1}
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
(b) Write the general solution.
(c) Solve equation (1) with the initial condition $u(x,0)= x$. Explain why the solution is fully determined by the initial condition.
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Characteristic Equation :$$\frac{dt}{1} = \frac{dx}{3(t^2-1)} = \frac{du}{6t^2} $$
a). From $$\frac{dt}{1} = \frac{dx}{3(t^2-1)} \\ (3t^2-3)dt = dx \\ \Rightarrow t^3 -3t +D =x \\ \Rightarrow D = x -t^3 -3t $$ b).From :$$\frac{dt}{1} = \frac{du}{6t^2} \\ 6t^2dt = du \\ \Rightarrow 2t^3+A = u\\ \Rightarrow u(x,t) = 2t^3+\phi(x-t^3-3t)$$ c). $$ u(x,0) = \phi(x) = x \\ \Rightarrow u(x,t) = 2t^3+x-t^3-3t \\ \Rightarrow u(x,t) = t^3+x-3t$$
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Jilong
In the last line of (a) should be $D=x-t^3 + 3t$. Correct from here
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