Question:
The solution of the differential equation $(x + 2y)dx + (2x - y)dy = 0$ is
The solution of the differential equation $(x + 2y)dx + (2x - y)dy = 0$ is
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If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, equation is exact.
Updated On: Apr 24, 2026
- $x^2 - y^2 + 6xy = C$
- $x^2 - y^2 - 4xy = C$
- $x^2 - y^2 + 4xy = C$
- $x^2 - y^2 + 3xy = C$
- $2x^2 - y^2 + 4xy = C$
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The Correct Option is C
Solution and Explanation
Concept:
• Check exact differential equation: $Mdx + Ndy = 0$
Step 1: Identify
\[ M = x + 2y,\quad N = 2x - y \]
Step 2: Check exactness
\[ \frac{\partial M}{\partial y} = 2,\quad \frac{\partial N}{\partial x} = 2 \] Hence exact.
Step 3: Integrate $M$ w.r.t $x$
\[ \int (x + 2y) dx = \frac{x^2}{2} + 2xy + \phi(y) \]
Step 4: Differentiate w.r.t $y$
\[ \frac{\partial}{\partial y} = 2x + \phi'(y) \] Equate with $N = 2x - y$: \[ \phi'(y) = -y \Rightarrow \phi(y) = -\frac{y^2}{2} \]
Step 5: Final solution
\[ \frac{x^2}{2} + 2xy - \frac{y^2}{2} = C \] Multiply by 2: \[ x^2 - y^2 + 4xy = C \] Final Conclusion:
Option (C)
• Check exact differential equation: $Mdx + Ndy = 0$
Step 1: Identify
\[ M = x + 2y,\quad N = 2x - y \]
Step 2: Check exactness
\[ \frac{\partial M}{\partial y} = 2,\quad \frac{\partial N}{\partial x} = 2 \] Hence exact.
Step 3: Integrate $M$ w.r.t $x$
\[ \int (x + 2y) dx = \frac{x^2}{2} + 2xy + \phi(y) \]
Step 4: Differentiate w.r.t $y$
\[ \frac{\partial}{\partial y} = 2x + \phi'(y) \] Equate with $N = 2x - y$: \[ \phi'(y) = -y \Rightarrow \phi(y) = -\frac{y^2}{2} \]
Step 5: Final solution
\[ \frac{x^2}{2} + 2xy - \frac{y^2}{2} = C \] Multiply by 2: \[ x^2 - y^2 + 4xy = C \] Final Conclusion:
Option (C)
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