To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is,

\(\displaystyle \\y'+ay=0 \\y(x)=c_1\cos(\sqrt{a}x)+c_2\sin(\sqrt{a}x)\)

Therefore, the equation becomes

\(\displaystyle y(x)=c_1\cos(4x)+c_2\sin(4x)\)

From here, apply the boundary conditions to solve for the constants \(\displaystyle c_1\) and \(\displaystyle c_2\)

\(\displaystyle \\-4=y(0)=c_1 \\\\5=y\left(\frac{\pi}{2} \right )=c_2\)

Thus resulting in the solution,

\(\displaystyle y(x)=-4\cos(4x)+5\sin(4x)\)

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is,

\(\displaystyle \\y'+ay=0 \\y(x)=c_1\cos(\sqrt{a}x)+c_2\sin(\sqrt{a}x)\)

Therefore, the equation becomes

\(\displaystyle y(x)=c_1\cos(3x)+c_2\sin(3x)\)

From here, apply the boundary conditions to solve for the constants \(\displaystyle c_1\) and \(\displaystyle c_2\)

\(\displaystyle \\-7=y(0)=c_1 \\\\2=y\left(\frac{\pi}{2} \right )=c_2\)

Thus resulting in the solution,

\(\displaystyle y(x)=-7\cos(3x)+2\sin(3x)\)

This statement is true by the Uniqueness Theorem.

\(\displaystyle \\u_{tt}=c^2u_{xx}, 0< x< 1, t>0 \\\text{Initial Conditions: }u(x,0)=f(x), u_t(x,0)=g(x), 0\leq x\leq l \\\text{Boundary Conditions: }u(0,t)=0, u(l,t)=0, t\geq 0\)

\(\displaystyle u(x,t)\) is twice differential equation in terms of \(\displaystyle x\) and \(\displaystyle t\).

Now, consider the energy integral

\(\displaystyle E(t)=\frac{1}{2}\int_0^l(c^2u^2_x+u^2_t)\ dx\)

After performing integration by parts results in the following,

\(\displaystyle \int_0^lc^2u_xu_{xt}\ dx=[c^2u_xu_t]_0^l-\int_0^lc^2u_tu_{xx}\ dx\)

From here, the initial conditions and boundary conditions are applied.

\(\displaystyle \\\frac{dE}{dt}=\int_0^l u_t(u_{tt}-c^2u_{xx})\ dx \\\\\frac{dE}{dt}=0 \\\\E(t)=C\)

Therefore,

\(\displaystyle u_1(x,t)=u_2(x,t)=u(x,t)\)

which proves the wave equations has only one solution and thus is unique.

The conservation law written as a partial differential equation is found by applying the divergence theorem to the conservation equation.

The conservation equation is,

\(\displaystyle \frac{d}{dt}\int_\wp U\ dx +\int_{\partial \wp} F(U)n\ ds=\int_\wp S(U,t)\ dx\)

Now, recall the divergence theorem which states,

\(\displaystyle \int_{\partial \wp}Fn\ ds=\int_{\wp}\bigtriangledown F\ dx\)

Thus, by substituting 

\(\displaystyle \int_{\partial \wp} Fn\ ds\) for \(\displaystyle \int_{\partial \wp} F(U)n\ ds\) results in,

\(\displaystyle \frac{d}{dt}\int_\wp U\ dx+\int_{\partial \wp}Fn\ ds=\int_\wp S\ dx\)

From here, rewriting this equation to bring the derivative inside the integral along with substituting

\(\displaystyle \int_{\partial \wp} Fn\ ds=\int_\wp \bigtriangledown F\ ds\),

and performing some algebraic operations results in,

\(\displaystyle \int_\wp \left(\frac{\partial U}{\partial t}+\bigtriangledown F-S \right )dx=0\)

After integrating over the domain \(\displaystyle \wp\) the partial differential equation that is found is,

\(\displaystyle \frac{\partial U}{\partial t}+\bigtriangledown F=S\)

Just like with ordinary differential equations, partial differential equations can be characterized by their order.

The order of an equation is defined by the highest ordered partial derivatives in the equations.

Looking at the equation in question,

\(\displaystyle u_{xx}+2u_{xy}+\frac{7}{5}u_{yy}=10y\)

The partial derivatives are:

\(\displaystyle \\u_{xx} \\u_{xy} \\u_{yy}\)

Notice that each partial derivative contains two variables, thus this equation is a second order partial differential equation.

Just like with ordinary differential equations, partial differential equations can be characterized by their order.

The order of an equation is defined by the highest ordered partial derivatives in the equations.

Looking at the equation in question,

\(\displaystyle u_{xyz}+u_{xy}+\frac{10}{3}=10y\)

The partial derivatives are:

\(\displaystyle \\u_{xyz} \\u_{xy}\)

Notice that one of them partial derivative contains three variables, thus this equation is a third order partial differential equation.

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.

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

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

\(\displaystyle u_{xx}+u_{yy}+u_{zz}=0\) is known as the Laplace equation.

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

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

Therefore, the correct answer for the biharmonic wave equation is 

\(\displaystyle u_{tt}+c^2\bigtriangledown^4u=0\)

Just like with ordinary differential equations, partial differential equations can be characterized by their order.

The order of an equation is defined by the highest ordered partial derivatives in the equations.

Looking at the equation in question,

\(\displaystyle u_{xyz}+u_{xy}=e^y\)

The partial derivatives are:

\(\displaystyle \\u_{xyz} \\u_{xy}\)

Notice that one of them partial derivative contains three variables, thus this equation is a third order partial differential equation.