On solving $\frac{d y}{d x}=\frac{x-y+3}{2 x-2 y+5}$, the solution obtained is $x=2(x-y)+\log (t)+c$, find $t$

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Aicharya · Accepted Answer

Given differential equation, $\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}$ Let $\quad x-y=t \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$, so $\begin{array}{ll} & 1-\frac{d t}{d x}=\frac{t+3}{2 t+5} \Rightarrow \frac{d t}{d x}=\frac{2 t+5-t-3}{2 t+5}=\frac{t+2}{2 t+5} \ \Rightarrow & \int \frac{2 t+5}{t+2} d t=\int d x \ \Rightarrow & \int\left(2+\frac{1}{t+2}\right) d t=x+C^{\prime} \ \Rightarrow & 2 t+\log _e|t+2|=x+C^{\prime} \ \Rightarrow & 2(x-y)+\log |x-y+2|+C=x \ \Rightarrow & 2(x-y)+\log (t)+C=x \quad \text { (given) } \end{array}$ On comparing, we get $\therefore \quad t=x-y+2$ Hence, option (a) is correct.

On solving \(\frac{d y}{d x}=\frac{x-y+3}{2 x-2 y+5}\), the solution obtained is \(x=2(x-y)+\log (t)+c\),…

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