TS EAMCET · Maths · Differential Equations
The general solution of the differential equat \(\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y\) is where \(c\) is an arbitrary constant
- A \(\mathrm{c}^2\left(x^2+y^2\right)=\left(y^2-x^2\right)\)
- B \(c^2\left(x^2+y^2\right)=\left(y^2-x^2\right)^2\)
- C \(c^2\left(x^2+y^2\right)^2=\left(y^2-x^2\right)\)
- D \(c^2\left(x^2-y^2\right)^2=\left(y^2-x^2\right)\)
Answer & Solution
Correct Answer
(C) \(c^2\left(x^2+y^2\right)^2=\left(y^2-x^2\right)\)
Step-by-step Solution
Detailed explanation
\(\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y\) \[ \Rightarrow \quad \frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y} \] It is homogeneous differential equation So, we have to take \(y=v x\) So, \[ \frac{d y}{d x}=v+x \frac{d v}{d x} \] Now,…
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