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Cards in this deck (17)
- 1\[ u'' + u = 0 \]Method: characteristic equation
Solution: \(u(x)=c_1 \cos x + c_2 \sin x\) - 2\[ u''+ 2u' + 3 = 0\]Method: characteristic equation
\(r^2 + 2r+3=0\) has roots \(r_{1,2} = -1 \pm i\sqrt{2}\)
Solution: \(u(x) = e^{-x}\left(c_1 \cos \sqrt{2}x + c_2 \sin \sqrt{2}x\right) \) - 3\[ u'' + u = f(x) \]Method: Find solution to homogeneous equation and particular solution for nonhomogeneous equation.
- 4\( a u_x + b u_y = 0 \) with \(a,b\) constantsMethod: Characteristic Curves
- 5\[ a(x,y)u_x + b(x,y)u_y = c(x,y) \]Method: Characteristic Curves.
- 6\(A u_{xx} + Bu_{xy} + Cu_{yy} = 0 \) with \(A,B,C\) real constants not all zeroMethod: Change of coordinates
\( \begin{cases}
\xi = \alpha x+ \beta y \\
\eta = \gamma x+ \delta y \\
\end{cases} \)
with \(\alpha \delta - \gamma \beta \neq 0\)
Solution: \(u(x,y) = F( \alpha x + \beta y)+G(\gamma x + \delta y) \) - 7Wave Equation \( u_{tt} - c^2 u_{xx}=0 \)Method: change of variables
\(\begin{cases}
\xi = x + ct \\
\eta = x-ct \\
\end{cases} \)
Solution: \(u(x,t) =F(x+ct) + G(x-ct) \) - 8Heat Equation \( u_t = u_{xx} \)Method: separation of variables
- 9Laplace Equation \( u_{xx} + u_{yy} = 0\)Method: polar coordinates, separation of variables.
- 10\( u'=u^2\) \(u(0)=\gamma\)Solution: \(\frac{\gamma}{1-\gamma t} \)
for \(t < \frac{1}{\gamma} \) - 11\[ u''+u = e^t\]Solution to homogeneous ODE: \( u(t) = c_1 \cos t + c_2 \sin t\)
Solution to nonhomogeneous ODE: \(u_p(t) = \frac{1}{2}e^t\)
General solution: \( u(t) = c_1 \cos t + c_2 \sin t + \frac{1}{2}e^t \) - 12\( u'' + u = 0 \) \(0 < t < \pi \) \(u(0)=0\) \(u(\pi)=1\)Method: Characteristic equation
Solution: No solution - 13\( u'' + u = 0 \) \(0 < t < \pi\) \(u(0)=u(\pi)=0\)Method: characteristic equation
Solution:
\(u(t) = c_1 \cos t\) - 14\[u'(t) + p(t) u(t) = q(t) \]Method: Multiply by integrating factor
\( \rho = e^{\int p dt} \)
Solution: Pick \(Q\) such that \(Q'=\rho q\).
\( u(t) = \frac{Q(t) + C}{\rho(t)} \) - 15Cauchy-Euler Type Equation \(\mathcal{L}u = \sum_{j=0}^n (t-t_0)^ja_j u^{(j)}=0\)Solutions of the form
\(u=(t-t_0)^\lambda\) with \( \lambda \) to be found. - 16\[ \int_0^1 u(y) dy = g(x) \] \[ 0 < x < 1\]Solution: \(g(x)=C\) for a constant \(C\). Infinitely many solutions.
- 17\[ \int_0^1 u(y) dy =\lambda u(x) + g(x) \] \[ 0 < x < 1\]Method: The solution must have the form \( u(x) = (C-g(x))/\lambda)\) for a constant \(C\). Plug into equation and solve for \(C\).
Solution: for continuous \(g\) and \(\lambda \neq 0,1\) the solution is
\(u(x) = \frac{\int_0^1 g(y)dy}{\lambda(1-\lambda)} -\frac{g(x)}{\lambda}\)
for \(\lambda =1\), \(u(x) = C-g(x)\)
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- Graduate Math 519 - Methods of Applied Math Flashcards on Problems, created by srein @iastate.edu on 25/09/2014.
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- srein @iastate.edu
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