TS EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}=\frac{2 x-3 y+4}{3 x+2 y-7}\) is
- A \(x^2+y^2=3 x y+y+C\)
- B \((2 x-3 y)^2+(3 x+2 y)^2=C\)
- C \(x^2+y^2+3 x y-4 x-7 y+C=0\)
- D \(x^2-3 x v-v^2+4 x+7 v+C=0\)
Answer & Solution
Correct Answer
(D) \(x^2-3 x v-v^2+4 x+7 v+C=0\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { We have, } \frac{d y}{d x}=\frac{2 x-3 y+4}{3 x+2 y-7} \\ & \Rightarrow \quad 3 x d y+2 y d y-7 d y=2 x d x-3 y d x+4 d x \\ & \Rightarrow \quad 3 x d y+3 y d x-7 d y+2 y d y-2 x d x-4 d x=0 \\ & \Rightarrow \quad 3 d(x y)+(2 y-7) d y-(4+2 x) d x=0…
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