Question

For the exact differential equation, \[ \frac{du}{dx}=\frac{-xu^{2}}{2+x^{2}u}, \] which one of the following is the solution?

• A. $u^{2}+2x^{2}=\text{constant}$
• B. $xu^{2}+u=\text{constant}$
• C. $\frac{1}{2}x^{2}u^{2}+2u=\text{constant}$
• D. $\frac{1}{2}ux^{2}+2x=\text{constant}$

Hint:

For exact differential equations, verify exactness by matching mixed partial derivatives, then integrate $M$ and adjust using $N$ to find the potential function.

Answer 1

The Correct Option is C

We are given the differential equation \[ \frac{du}{dx}=\frac{-xu^{2}}{2+x^{2}u}. \] We rearrange it to check if it is exact. Write it as \[ (2+x^{2}u)\,du + xu^{2}\,dx = 0. \] This takes the form \[ M(x,u)\,dx + N(x,u)\,du = 0, \] where \[ M = xu^{2}, \qquad N = 2 + x^{2}u. \] Step 1: Check exactness.
Compute partial derivatives: \[ \frac{\partial M}{\partial u} = 2xu, \qquad \frac{\partial N}{\partial x} = 2xu. \] Since both are equal, the equation is exact.
Step 2: Integrate $M$ with respect to $x$.
\[ \int xu^{2}\,dx = \frac{1}{2}x^{2}u^{2} + f(u). \] Step 3: Differentiate with respect to $u$.
\[ \frac{\partial}{\partial u}\left(\frac{1}{2}x^{2}u^{2} + f(u)\right) = x^{2}u + f'(u). \] This must equal $N = 2 + x^{2}u$, therefore \[ f'(u) = 2 \quad \Rightarrow \quad f(u)=2u. \] Step 4: Write the solution.
\[ \frac{1}{2}x^{2}u^{2} + 2u = C. \] Final Answer: $\frac{1}{2}x^{2}u^{2}+2u=\text{constant}$