MHT CET · Maths · Differential Equations
The solution of the differential equation \(\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}\) is
- A \(2(x-y)+\log (x-y)=x+c\)
- B \(2(x-y)-\log (x-y+2)=x+c\)
- C \(2(x-y)+\log (x-y+2)=x+c\)
- D None of the above
Answer & Solution
Correct Answer
(C) \(2(x-y)+\log (x-y+2)=x+c\)
Step-by-step Solution
Detailed explanation
Given differential equation is
\(
\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}
\)
Let \(\quad x-y=v \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x}\)
\(\therefore\) \(
\frac{d v}{d x}=\frac{v+2}{2 v+5}
\)
\(\Rightarrow \int \frac{2 v+5}{v+2} d v=\int d x \)
\( \Rightarrow \int\left(2+\frac{1}{v+2}\right) d v=\int d x\)
\(\Rightarrow 2 v+\log (v+2)=x+c\)
\(\Rightarrow 2(x-y)+\log (x-y+2)=x+c\)
\(
\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}
\)
Let \(\quad x-y=v \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x}\)
\(\therefore\) \(
\frac{d v}{d x}=\frac{v+2}{2 v+5}
\)
\(\Rightarrow \int \frac{2 v+5}{v+2} d v=\int d x \)
\( \Rightarrow \int\left(2+\frac{1}{v+2}\right) d v=\int d x\)
\(\Rightarrow 2 v+\log (v+2)=x+c\)
\(\Rightarrow 2(x-y)+\log (x-y+2)=x+c\)
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