Question

The equation of the curve passing through (3, 4) and satisfying the differential equation $y\left(\frac{dy}{dx}\right)^2+(x-y)\frac{dy}{dx}-x=0$ can be:

• A. x-y+1=0
• B. x^2+y^2 = 25
• C. x^2+y^2-5x-10=0
• D. x+y-7 = 0

Answer 1

Answer: (A), (B)

x-y+1=0, x^2+y^2 = 25

Answer 2

The given differential equation is $y\left(\frac{dy}{dx}\right)^2+(x-y)\frac{dy}{dx}-x=0$. Let $p = \frac{dy}{dx}$. The equation becomes $yp^2 + (x-y)p - x = 0$. Factoring this quadratic in $p$, we get $yp^2 + xp - yp - x = 0$, which simplifies to $p(yp+x) - 1(yp+x) = 0$, or $(p-1)(yp+x) = 0$.

This yields two cases:

Case 1: $p - 1 = 0 \implies \frac{dy}{dx} = 1$. Integrating, we get $y = x + C_1$. Since the curve passes through (3, 4), we substitute these values: $4 = 3 + C_1$, which gives $C_1 = 1$. Thus, the equation of the curve is $y = x + 1$, or $x - y + 1 = 0$.

Case 2: $yp + x = 0 \implies yp = -x \implies \frac{dy}{dx} = -\frac{x}{y}$. This is a separable differential equation: $y dy = -x dx$. Integrating both sides, we get $\int y dy = \int -x dx$, which results in $\frac{y^2}{2} = -\frac{x^2}{2} + C_2$. Multiplying by 2, we get $y^2 = -x^2 + 2C_2$, or $x^2 + y^2 = 2C_2$. Let $K = 2C_2$. So, the equation is $x^2 + y^2 = K$. Since the curve passes through (3, 4), we substitute these values: $3^2 + 4^2 = K$, which gives $9 + 16 = K$, so $K = 25$. Thus, the equation of the curve is $x^2 + y^2 = 25$.

Both $x-y+1=0$ and $x^2+y^2=25$ are valid equations for curves passing through (3, 4) and satisfying the given differential equation.