AP EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}+\frac{x+y+1}{x-3 y+5}=0\) is
- A \(3(\mathrm{y}-1)^2-2(\mathrm{x}+2)(\mathrm{y}-1)-(\mathrm{x}+2)^2=\mathrm{c}\)
- B \(x^2-3 y^2-4 x y-2 x-10 y=c\)
- C \(3(\mathrm{y}+1)^2+2(\mathrm{x}-2)(\mathrm{y}+1)-(\mathrm{x}-2)^2=\mathrm{c}\)
- D \(x^2+3 y^2+4 x y+2 x+10 y=c\)
Answer & Solution
Correct Answer
(A) \(3(\mathrm{y}-1)^2-2(\mathrm{x}+2)(\mathrm{y}-1)-(\mathrm{x}+2)^2=\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\(x+y+1=0, x-3y+5=0 \Rightarrow y=1, x=-2\) Let \(x=X-2, y=Y+1\). Then \(\frac{dY}{dX} = -\frac{X+Y}{X-3Y}\). Let \(Y=vX \Rightarrow v+X\frac{dv}{dX} = -\frac{1+v}{1-3v}\). \(X\frac{dv}{dX} = \frac{-1-v-v+3v^2}{1-3v} = \frac{3v^2-2v-1}{1-3v}\).…
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