Question

If $(2xy - y^2 - y)dx = (2xy + x - x^2)dy$ and $y(1) = 1$, then the value of $12|y(-1)|$ is

Answer 1

12

Answer 2

The given differential equation is $(2xy - y^2 - y)dx = (2xy + x - x^2)dy$. We can rewrite this as $(2xy - y^2 - y)dx - (2xy + x - x^2)dy = 0$. Let $M = 2xy - y^2 - y$ and $N = -(2xy + x - x^2) = x^2 - 2xy - x$. We check if the differential equation is exact by verifying if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy - y^2 - y) = 2x - 2y - 1$. $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 - 2xy - x) = 2x - 2y - 1$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact.

For an exact differential equation $M dx + N dy = 0$, there exists a function $F(x, y)$ such that $dF = M dx + N dy$. This means $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$. We integrate $M$ with respect to $x$ to find $F(x, y)$: $F(x, y) = \int M dx = \int (2xy - y^2 - y) dx = x^2y - xy^2 - xy + g(y)$, where $g(y)$ is an arbitrary function of $y$.

Now, we differentiate $F(x, y)$ with respect to $y$ and set it equal to $N$: $\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(x^2y - xy^2 - xy + g(y)) = x^2 - 2xy - x + g'(y)$. We have $N = x^2 - 2xy - x$. So, $x^2 - 2xy - x + g'(y) = x^2 - 2xy - x$. This implies $g'(y) = 0$. Integrating with respect to $y$, we get $g(y) = C_1$, where $C_1$ is a constant.

The general solution is $F(x, y) = C$, which is $x^2y - xy^2 - xy + C_1 = C_2$. Let $C = C_2 - C_1$. The general solution is $x^2y - xy^2 - xy = C$.

We are given the initial condition $y(1) = 1$. We substitute $x=1$ and $y=1$ into the general solution to find the value of $C$: $(1)^2(1) - (1)(1)^2 - (1)(1) = C$ $1 - 1 - 1 = C$ $C = -1$.

The particular solution is $x^2y - xy^2 - xy = -1$.

We need to find the value of $12|y(-1)|$. Let $y(-1) = y_0$. We substitute $x=-1$ into the particular solution: $(-1)^2 y_0 - (-1) y_0^2 - (-1) y_0 = -1$ $1 \cdot y_0 - (-1) y_0^2 - (-1) y_0 = -1$ $y_0 + y_0^2 + y_0 = -1$ $y_0^2 + 2y_0 + 1 = 0$ $(y_0 + 1)^2 = 0$ This gives $y_0 = -1$.

So, $y(-1) = -1$.

Finally, we need to find the value of $12|y(-1)|$: $12|y(-1)| = 12|-1| = 12 \times 1 = 12$.