TS EAMCET · Maths · Differential Equations
The solution of \(\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0\) is :
- A \(3 x\left(1+y^2\right)=4 y^3+c\)
- B \(3 y\left(1+x^2\right)=4 x^3+c\)
- C \(3 x\left(1-y^2\right)=4 y^3+c\)
- D \(3 y\left(1+y^2\right)=4 x^3+c\)
Answer & Solution
Correct Answer
(B) \(3 y\left(1+x^2\right)=4 x^3+c\)
Step-by-step Solution
Detailed explanation
\(\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0\) \(\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}\) On comparing with \(\frac{d y}{d x}+P y=Q\), we get \(P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2}\) I.F.…
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