Question

If \( u = x^3 \) and \( v = y^2 \) transform the differential equation \( 3x^5 dx - y(y^2 - x^3) dy = 0 \) to \( \dfrac{dv}{du} = \dfrac{\alpha u}{2(u - v)} \), then \( \alpha \) is
 

• A. 4
• B. 2
• C. -2
• D. -4

Hint:

Always convert \( dx, dy \) correctly using \( du, dv \) and simplify before substituting back.

Answer 1

The Correct Option is D

Step 1: Find the Differentials

We have the transformation equations:

$$u = x^3 \implies du = \frac{d}{dx}(x^3) dx \implies du = 3x^2 dx$$

From this, we get an expression for $dx$:

$$dx = \frac{du}{3x^2}$$

$$v = y^2 \implies dv = \frac{d}{dy}(y^2) dy \implies dv = 2y dy$$

From this, we get an expression for $dy$:

$$dy = \frac{dv}{2y}$$

Step 2: Substitute into the Differential Equation

Substitute the expressions for $dx$, $dy$, $u$, and $v$ into the original differential equation:

$$3x^5 dx - y(y^2 - x^3) dy = 0$$

$$3x^5 \left( \frac{du}{3x^2} \right) - y(v - u) \left( \frac{dv}{2y} \right) = 0$$

Step 3: Simplify the Equation

Simplify the terms:

$$\left( \frac{3x^5}{3x^2} \right) du - \left( \frac{y}{2y} \right) (v - u) dv = 0$$

$$x^3 du - \frac{1}{2} (v - u) dv = 0$$

Now substitute $u = x^3$ back into the simplified equation:

$$u du - \frac{1}{2} (v - u) dv = 0$$

Multiply by 2 to clear the fraction:

$$2u du - (v - u) dv = 0$$

Rearrange the terms to isolate the $dv$ term:

$$(v - u) dv = 2u du$$

Step 4: Find the Ratio $\frac{dv}{du}$ and Determine $\alpha$

Divide by $(v - u) du$ to express $\frac{dv}{du}$:

$$\frac{dv}{du} = \frac{2u}{v - u}$$

We are given the target form $\frac{dv}{du} = \frac{\alpha u}{2(u - v)}$. To match our result, we manipulate the denominator of our result:

$$\frac{dv}{du} = \frac{2u}{-(u - v)}$$

$$\frac{dv}{du} = \frac{-2u}{u - v}$$

To match the target form, we multiply the numerator and denominator by 2:

$$\frac{dv}{du} = \frac{2 \cdot (-2) u}{2(u - v)} = \frac{-4u}{2(u - v)}$$

Comparing this result to the required form $\frac{dv}{du} = \frac{\alpha u}{2(u - v)}$, we find:

$$\alpha = -4$$

The correct value is $\alpha = -4$.