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  • The solution of equation (2 y-1) d x-(2 x+3) d y=0 is (a) frac{2 x-1}{2 y+3}=k

The solution of equation (2 y-1) d x-(2 x+3) d y=0 is
(a) \frac{2 x-1}{2 y+3}=k
(b) \frac{2 y+1}{2 x-3}=k
(c) \frac{2 x+3}{2 y-1}=k
(d)  \frac{2 x-1}{2 y-1}=k

Answers (1)

The answer is the option (c)

Explanation: -

\begin{aligned} &\begin{array}{ll} \Rightarrow & (2 y-1) d x=(2 x+3) d y \\ \Rightarrow & \frac{d x}{2 x+3}=\frac{d y}{2 y-1} \end{array}\\ &\text { On integrating both sides, we get }\\ &\frac{1}{2} \log (2 x+3)=\frac{1}{2} \log (2 y-1)+\log C\\ &\begin{array}{ll} \Rightarrow & {[\log (2 x+3)-\log (2 y-1)]=2 \log C} \\ \Rightarrow & \log \left(\frac{2 x+3}{2 y-1}\right)=\log C^{2} \\ \Rightarrow & \frac{2 x+3}{2 y-1}=C^{2} \\ \Rightarrow & \frac{2 x+3}{2 y-1}=k, \text { where } K=C^{2} \end{array} \end{aligned}

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