MCQ
The solution of ${e^{2x - 3y}}dx + {e^{2y - 3x}}dy = 0$ is
- ✓${e^{5x}} + {e^{5y}} = c$
- B${e^{5x}} - {e^{5y}} = c$
- C${e^{5x + 5y}} = c$
- DNone of these
Multiply the equation by ${e^{3x + 3y}}$ ==> ${e^{5x}}dx + {e^{5y}}dy = 0$
On integrating, we get ${e^{5x}} + {e^{5y}} = 5c' = c$.
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$\frac{d y}{d x}=x y-1+x-y ; y(0)=0$