Question

An integrating factor of the differential equation 
\[ \left( y + \frac{1}{3} y^3 + \frac{1}{2} x^2 \right) \, dx + \frac{1}{4} (x + xy^2) \, dy = 0 \] is 
 

• A. \( x^2 \)
• B. \( 3 \log_e x \)
• C. \( x^3 \)
• D. \( 2 \log_e x \)

Hint:

For non-exact differential equations, the integrating factor often depends on the variable that causes the equation to become exact.

Answer 1

The Correct Option is C

Step 1: Check if the equation is exact

For exactness, we need $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

$$\frac{\partial M}{\partial y} = 1 + y^2$$

$$\frac{\partial N}{\partial x} = \frac{1}{4}(1 + y^2)$$

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.

Step 2: Find the integrating factor

Let's check if there's an integrating factor that depends only on $x$, say $\mu(x)$.

For an integrating factor $\mu(x)$: $$\frac{d\mu}{dx} = \mu \cdot \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$$

Calculate: $$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(1 + y^2) - \frac{1}{4}(1 + y^2)}{\frac{1}{4}x(1 + y^2)}$$

$$= \frac{\frac{3}{4}(1 + y^2)}{\frac{1}{4}x(1 + y^2)} = \frac{3}{x}$$

So: $$\frac{d\mu}{\mu} = \frac{3}{x}dx$$

Integrating: $$\ln|\mu| = 3\ln|x| + C$$

$$\mu = Kx^3$$

Taking $K = 1$, we get $\mu(x) = x^3$

Step 3: Verify that $x^3$ is an integrating factor

Multiply the equation by $x^3$: $$x^3\left(y + \frac{1}{3}y^3 + \frac{1}{2}x^2\right)dx + x^3 \cdot \frac{1}{4}(x + xy^2)dy = 0$$

$$\left(x^3y + \frac{1}{3}x^3y^3 + \frac{1}{2}x^5\right)dx + \frac{1}{4}(x^4 + x^4y^2)dy = 0$$

Now:

• $M_1 = x^3y + \frac{1}{3}x^3y^3 + \frac{1}{2}x^5$
• $N_1 = \frac{1}{4}x^4(1 + y^2)$

Check exactness: $$\frac{\partial M_1}{\partial y} = x^3 + x^3y^2 = x^3(1 + y^2)$$

$$\frac{\partial N_1}{\partial x} = \frac{1}{4} \cdot 4x^3(1 + y^2) = x^3(1 + y^2)$$

Since $\frac{\partial M_1}{\partial y} = \frac{\partial N_1}{\partial x}$, the equation is now exact

Answer: (C)