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Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise  Revision Exercise Question 39 Maths Textbook Solution.

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2\left ( x+y \right )-4x-log|2x+2y-1|=4C

Hint: you must know the rules of solving differential equation and integrations.

Given:\left ( x+y-1 \right )dy=\left ( x+y \right )dx


\left ( x+y-1 \right )dy=\left ( x+y \right )dx

\frac{dy}{dx}=\frac{\left ( x+y \right )}{x+y-1}

Putting x+y=v , we get,

\begin{aligned} &1+\frac{d y}{d x}=\frac{d v}{d x} \\ &\frac{d y}{d x}=\frac{d v}{d x}-1 \\ &\frac{d v}{d x}-1=\frac{v}{(v-1)} \\ &\frac{d v}{d x}=\frac{v}{(v-1)}+1 \\ &\frac{d v}{d x}=\frac{v+v-1}{(v-1)} \\ &\frac{d v}{d x}=\frac{2 v-1}{(v-1)} \\ &\frac{v-1}{2 v-1} d v=d x \end{aligned}

Integration both sides we get

\begin{aligned} &\int \frac{v-1}{2 v+1} d v=\int d x \\ &\frac{1}{2} \int \frac{2 v}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int \frac{2 v-1+1}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int d v+\frac{1}{2} \int \frac{1}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int d v-\frac{1}{2} \int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} v-\frac{1}{4} \log |2 v-1|=x+C \end{aligned}

\begin{aligned} &\frac{1}{2}(x+y)-\frac{1}{4} \log |2 x+2 y-1|=x+C \\ &2(x+y)-\log |2 x+2 y-1|=4 x+4 C \\ &2(x+y)-4 x-\log |2 x+2 y-1|=4 C \end{aligned}


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