Question

Consider a curve passing through the points (0, 0) and ($x_1$, 2), whose differential equation is $(x + y - 2) dy + (y - x) dx = 0$. Then the value of $x_1$ is

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (D)

2

Answer 2

The given differential equation is $(x + y - 2) dy + (y - x) dx = 0$. This can be rewritten as $\frac{dy}{dx} = \frac{x - y}{x + y - 2}$. This is a non-homogeneous linear differential equation. We can make it homogeneous by a substitution. Let $x = X + h$ and $y = Y + k$. Then $dx = dX$ and $dy = dY$. Substituting these into the equation: $((X+h) + (Y+k) - 2) dY + ((Y+k) - (X+h)) dX = 0$ $(X + Y + h + k - 2) dY + (Y - X + k - h) dX = 0$ To make this homogeneous, we need the constant terms to be zero:

1. $h + k - 2 = 0$
2. $k - h = 0$ From (2), $k = h$. Substituting into (1): $h + h - 2 = 0 \implies 2h = 2 \implies h = 1$. Thus, $k = 1$. The substitution is $x = X + 1$ and $y = Y + 1$, which means $X = x - 1$ and $Y = y - 1$. The differential equation in terms of $X$ and $Y$ becomes: $(X + Y) dY + (Y - X) dX = 0$ $\frac{dY}{dX} = \frac{X - Y}{X + Y}$ This is a homogeneous differential equation. Let $Y = vX$. Then $\frac{dY}{dX} = v + X\frac{dv}{dX}$. $v + X\frac{dv}{dX} = \frac{X - vX}{X + vX} = \frac{1 - v}{1 + v}$ $X\frac{dv}{dX} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v(1 + v)}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$ Separating variables: $\frac{1 + v}{1 - 2v - v^2} dv = \frac{dX}{X}$ Integrating both sides: $\int \frac{1 + v}{1 - 2v - v^2} dv = \int \frac{dX}{X}$ The integral on the left is $-\frac{1}{2} \ln|1 - 2v - v^2|$. So, $-\frac{1}{2} \ln|1 - 2v - v^2| = \ln|X| + C_1$ $\ln|1 - 2v - v^2|^{-1/2} = \ln|X| + C_1$ $\frac{1}{\sqrt{1 - 2v - v^2}} = A|X|$ where $A = e^{C_1}$ Squaring both sides: $\frac{1}{1 - 2v - v^2} = A^2 X^2$ $1 = A^2 X^2 (1 - 2\frac{Y}{X} - \frac{Y^2}{X^2})$ $1 = A^2 (X^2 - 2XY - Y^2)$ $X^2 - 2XY - Y^2 = C'$ where $C' = 1/A^2$. Substituting back $X = x - 1$ and $Y = y - 1$: $(x - 1)^2 - 2(x - 1)(y - 1) - (y - 1)^2 = C'$ The curve passes through (0, 0). Substitute $x=0, y=0$: $(-1)^2 - 2(-1)(-1) - (-1)^2 = C'$ $1 - 2 - 1 = C' \implies C' = -2$. The equation of the curve is $(x - 1)^2 - 2(x - 1)(y - 1) - (y - 1)^2 = -2$. The curve passes through ($x_1$, 2). Substitute $x=x_1, y=2$: $(x_1 - 1)^2 - 2(x_1 - 1)(2 - 1) - (2 - 1)^2 = -2$ $(x_1 - 1)^2 - 2(x_1 - 1)(1) - (1)^2 = -2$ $(x_1 - 1)^2 - 2(x_1 - 1) - 1 = -2$ Let $Z = x_1 - 1$. Then $Z^2 - 2Z - 1 = -2 \implies Z^2 - 2Z + 1 = 0$. This factors as $(Z - 1)^2 = 0$, so $Z = 1$. Substituting back $Z = x_1 - 1$, we get $x_1 - 1 = 1$, which means $x_1 = 2$.