The image contains a question prompt related to solving partial differential equations using the separation of variables method.**Question 1:**Use separation of variables to find, if possible, product solutions for the given partial differential equation. (Use the separation constant \(-\lambda \neq 0\). If not possible, enter IMPOSSIBLE.)\[\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y}\]\(u(x, y) = \)[An input box is provided here]A red "X" is displayed next to the input box, indicating an incorrect answer was submitted.**Question 2:**Below, another question is presented, labeled "ZILLDIFFEQ9 12.1.002."Use separation of variables to find, if possible, product solutions for the given partial differential equation. (Use the separation constant \(-\lambda \neq 0\). If not possible, enter IMPOSSIBLE.)\[\frac{\partial u}{\partial x} + 7 \frac{\partial u}{\partial y} = 0\]\(u(x, y) = c \cdot e^{\lambda x + \left(\frac{\lambda}{7} y\right)}\)Another red "X" is displayed, indicating the given answer is incorrect.Buttons are available for actions titled "DETAILS," "PREVIOUS ANSWERS," "MY NOTES," and "ASK YOUR TEACHER."This content introduces partial differential equations and hints at problem-solving through separation of variables, an important method used in mathematical physics and engineering.

Given∂u∂x=∂u∂y-------(1) Let u=X(x) Y(y) be the solution of equation Hence ∂[X.Y]∂x=∂[X.Y]∂y⇒X1.Y=X.Y1⇒X1X=Y1Y=-λ, since two expression in variables are equal when both are equal to a constant here -λ is positive or negative or zero Case 1) -λ=postive or negative[or]not equal to zero hence x1x=-λ ⇒dXdx=-λX⇒dXX=-λdx, which is variable separable formOn integration ; ln(x)=-λx+ln(A)⇒X=A.e-λxY1Y=-λ ⇒dYdy=-λY⇒dYY=-λdy, which is variable separable formOn integration ; ln(Y)=-λy+ln(B)⇒Y=B.e-λy Hence solution is u(x,y)=(A.e-λx)(B.e-λy)=AB.e-λ(x+y)Given∂u∂x+7∂u∂y=0-------(1) Let u=X(x) Y(y) be the solution of equation Hence ∂[X.Y]∂x=-7∂[X.Y]∂y⇒X1.Y=-7X.Y1⇒X1X=-7Y1Y=-λ, since two expression in variables are equal when both are equal to a constant here -λ is positive or negative or zero Case 1) -λ=postive or negative[or]not equal to zero hence x1x=-λ ⇒dXdx=-λX⇒dXX=-λdx, which is variable separable formOn integration ; ln(x)=-λx+ln(A)⇒X=A.e-λx-7Y1Y=-λ ⇒dYdy=λ7Y⇒dYY=λ7dy, which is variable separable formOn integration ; ln(Y)=λ7y+ln(B)⇒Y=B.eλ7y Hence solution is u(x,y)=(A.e-λx)(B.eλ7y)=AB.e-λ(x-y7)