\text{Consider the differential equation}\frac{\partial }{y}=3\frac{\partial }{y}+y $\text{If}\mathrm{ y}(x,0)=10e^{-2x}\text{, then find}y(x,t)$

The correct option is:
$y(x,t)=10e^{-2x-t}$

Step-by-step Derivation:

We are given the partial differential equation:
$$\frac{\partial y}{\partial x}=3\frac{\partial y}{\partial t}+y$$
subject to the initial condition:
$$y(x,0)=10e^{-2x}$$

We can solve this partial differential equation using the method of separation of variables. Let us assume a product solution of the form:
$$y(x,t)=X\left(x\right)T\left(t\right)$$

Taking the partial derivatives with respect to $x$ and $t$, we get:
$$\frac{\partial y}{\partial x}=X^{\prime }\left(x\right)T\left(t\right)$$
and
$$\frac{\partial y}{\partial t}=X\left(x\right)T^{\prime }\left(t\right)$$

Substituting these derivatives back into the original differential equation gives:
$$X^{\prime }\left(x\right)T\left(t\right)=3X\left(x\right)T^{\prime }\left(t\right)+X\left(x\right)T\left(t\right)$$

To separate the variables, we divide both sides of the equation by $X\left(x\right)T\left(t\right)$:
$$\frac{X^{\prime }\left(x\right)}{X\left(x\right)}=3\frac{T^{\prime }\left(t\right)}{T\left(t\right)}+1$$

Since the left-hand side is a function only of $x$ and the right-hand side is a function only of $t$, they must be equal to a separation constant, which we will call $k$:
$$\frac{X^{\prime }\left(x\right)}{X\left(x\right)}=k$$
and
$$3\frac{T^{\prime }\left(t\right)}{T\left(t\right)}+1=k$$

Now, we solve these two ordinary differential equations separately.

For the first equation involving $X\left(x\right)$:
$$\frac{dX}{dx}=kX\Rightarrow X\left(x\right)=A{e}^{kx}$$
where $A$ is a constant.

For the second equation involving $T\left(t\right)$:
$$3\frac{T^{\prime }\left(t\right)}{T\left(t\right)}=k-1\Rightarrow \frac{dT}{dt}=\left(\frac{k-1}{3}\right)T\Rightarrow T\left(t\right)=B{e}^{\left(\frac{k-1}{3}\right)t}$$
where $B$ is a constant.

Combining the solutions, the product solution is:
$$y(x,t)=C{e}^{kx+\left(\frac{k-1}{3}\right)t}$$
where $C=AB$ is a combined constant.

Now, we apply the initial condition $y(x,0)=10e^{-2x}$ by setting $t=0$:
$$y(x,0)=C{e}^{kx}=10e^{-2x}$$

Comparing the coefficients and the exponents on both sides, we find:
$$C=10$$
and
$$k=-2$$

Substituting these values back into the expression for $y(x,t)$, we obtain:
$$y(x,t)=10e^{-2x+\left(\frac{-2-1}{3}\right)t}$$

Simplifying the coefficient of $t$ in the exponent:
$$\frac{-2-1}{3}=\frac{-3}{3}=-1$$

This gives final expression for $y(x,t)$:
$$y(x,t)=10e^{-2x-t}$$