Second Order Linear PDEs - Differential Equations

Card 0 of 12

Question

Which of the following describes the physical phenomena that is the heat equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the heat equation is

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Solve for the general solution using separation of variables.

Given the following information:

$\u_t=u_{xx} \0\leq x\leq1 \t>0$

$\\text{Boundary Conditions: }\left{\begin{matrix} u_x(0,t)=0\ u_x(1,t)=0 \end{matrix}\right.$

$\text{Initial Conditions: }u(x,0)=cos(\pi x)$

Answer

Since the question states to use separation of variables the solution looks as follows.

$u(x,0)=cos(\pi x)$

Let

therefore the partial differential equation becomes

$\X(x)T'(t)=T(t)X''(x) \\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

is some constant therefore making the ordinary differential equation,

$\T'(t)-kT(t)=0 \X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\k=-\lambda^2 \T'(t)+\lambda^2T(t)=0 \X''(x)+\lambda^2X(x)=0$

Solving for and results in the following.

$\T(t)=Ae^{-\lambda^2t} \X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \B=0 \u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the wave equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the wave equation is

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the heat equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the heat equation is

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Solve for the general solution using separation of variables.

Given the following information:

$\u_t=u_{xx} \0\leq x\leq1 \t>0$

$\\text{Boundary Conditions: }\left{\begin{matrix} u_x(0,t)=0\ u_x(1,t)=0 \end{matrix}\right.$

$\text{Initial Conditions: }u(x,0)=cos(\pi x)$

Answer

Since the question states to use separation of variables the solution looks as follows.

$u(x,0)=cos(\pi x)$

Let

therefore the partial differential equation becomes

$\X(x)T'(t)=T(t)X''(x) \\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

is some constant therefore making the ordinary differential equation,

$\T'(t)-kT(t)=0 \X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\k=-\lambda^2 \T'(t)+\lambda^2T(t)=0 \X''(x)+\lambda^2X(x)=0$

Solving for and results in the following.

$\T(t)=Ae^{-\lambda^2t} \X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \B=0 \u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the wave equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the wave equation is

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the heat equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the heat equation is

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Solve for the general solution using separation of variables.

Given the following information:

$\u_t=u_{xx} \0\leq x\leq1 \t>0$

$\\text{Boundary Conditions: }\left{\begin{matrix} u_x(0,t)=0\ u_x(1,t)=0 \end{matrix}\right.$

$\text{Initial Conditions: }u(x,0)=cos(\pi x)$

Answer

Since the question states to use separation of variables the solution looks as follows.

$u(x,0)=cos(\pi x)$

Let

therefore the partial differential equation becomes

$\X(x)T'(t)=T(t)X''(x) \\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

is some constant therefore making the ordinary differential equation,

$\T'(t)-kT(t)=0 \X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\k=-\lambda^2 \T'(t)+\lambda^2T(t)=0 \X''(x)+\lambda^2X(x)=0$

Solving for and results in the following.

$\T(t)=Ae^{-\lambda^2t} \X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \B=0 \u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the wave equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the wave equation is

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the heat equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the heat equation is

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

Question

Solve for the general solution using separation of variables.

Given the following information:

$\u_t=u_{xx} \0\leq x\leq1 \t>0$

$\\text{Boundary Conditions: }\left{\begin{matrix} u_x(0,t)=0\ u_x(1,t)=0 \end{matrix}\right.$

$\text{Initial Conditions: }u(x,0)=cos(\pi x)$

Answer

Since the question states to use separation of variables the solution looks as follows.

$u(x,0)=cos(\pi x)$

Let

therefore the partial differential equation becomes

$\X(x)T'(t)=T(t)X''(x) \\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

is some constant therefore making the ordinary differential equation,

$\T'(t)-kT(t)=0 \X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\k=-\lambda^2 \T'(t)+\lambda^2T(t)=0 \X''(x)+\lambda^2X(x)=0$

Solving for and results in the following.

$\T(t)=Ae^{-\lambda^2t} \X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \B=0 \u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Compare your answer with the correct one above

Question

Which of the following describes the physical phenomena that is the wave equation?

Answer

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

$u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the wave equation is

$u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$

Compare your answer with the correct one above

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