Separable Equations — Question 5

Question 5 – Solution

Strategy. Test whether variables actually factor before performing separation; a sum of terms can still conceal a product.

Step 1: Establish the necessary product test. If $f(x,y)=g(x)h(y)$, then both products in the stated identity equal $g(a)g(b)h(c)h(d)$. This argument does not divide by any factor and remains valid if some factors vanish.

Step 2: Reject direct separation for (A). For $f(x,y)=x+y$, the proposed rectangle gives $$f(1,1)f(2,2)=2\cdot 4=8,\qquad f(1,2)f(2,1)=3\cdot 3=9.$$ The necessary identity fails, so there is no product representation on this rectangle. This conclusion concerns direct separation on that region; it makes no claim about other methods of solving (A).

Step 3: Solve the factored equation (B). Here $xy+x+y+1=(x+1)(y+1)$. First, $y\equiv-1$ is a constant solution. For $y\ne-1$, $$\frac{dy}{y+1}=(x+1)\,dx,\qquad \ln|y+1|=\frac{x^2}{2}+x+C.$$ Absorbing the fixed sign into a nonzero constant gives $$\boxed{y=-1+A e^{x^2/2+x}}.$$ Allowing $A=0$ includes the constant solution as well. These are all solutions: for any differentiable solution, the product rule applied to $(y+1)e^{-x^2/2-x}$ gives derivative zero, even at $y=-1$.

Step 4: Apply and check the data. The initial condition requires $A=-1$, so $\boxed{y=-1-e^{x^2/2+x}}$ on all of $\mathbb R$. Its derivative is $-(x+1)e^{x^2/2+x}=(x+1)(y+1)$, and $y(0)=-2$. Separation integrates the factor $x+1$; it never requires its reciprocal. The point $x=-1$ is an ordinary point with $y'=0$.