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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.

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

Answers (1)

Answer:

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

Solution:

\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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