Chapter 13: Problem 17
Identifying \(A=B=C=1,\) we compute \(B^{2}-4 A C=-3 < 0 .\) The equation is elliptic.
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Chapter 13: Problem 17
Identifying \(A=B=C=1,\) we compute \(B^{2}-4 A C=-3 < 0 .\) The equation is elliptic.
These are the key concepts you need to understand to accurately answer the question.
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Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(X^{\prime} Y=X Y^{\prime}+X Y .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=\frac{Y+Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad y^{\prime}+(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\\ &Y=c_{2} e^{-(1+\lambda) y}=0. \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-y-\lambda(x+y)}.$$
$$\begin{aligned}
&k \frac{\partial^{2} u}{\partial x^{2}}-h u=\frac{\partial u}{\partial t},
\quad 0
Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(y X^{\prime} Y^{\prime}+X Y=0 .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=-\frac{Y}{y Y^{\prime}}=-\lambda.$$ When \(\lambda \neq 0\) $$X^{\prime}+\lambda X=0 \quad \text { and } \quad \lambda y Y^{\prime}-Y=0$$ so that $$X=c_{1} e^{-\lambda x} \quad \text { and } \quad Y=c_{2} y^{1 / \lambda}.$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-\lambda x} y^{1 / \lambda}.$$ In this case \(\lambda=0\) yields no solution.
$$\begin{array}{l}
a^{2} \frac{\partial^{2} u}{\partial x^{2}}-2 \beta \frac{\partial u}{\partial
t}=\frac{\partial^{2} u}{\partial t^{2}}, \quad 0
$$\begin{array}{l}
a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial
t^{2}}, \quad 0
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