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MHT CET · Maths · Differential Equations

General solution of the differential equation \(\left(y^3+y\right)\left(x^2+1\right) d y=\left(x y^4+2 y^2 x\right) d x\) is (where \(C\) is a constant of integration.)

  1. A \(\mathrm{y}^2\left(\mathrm{y}^2+1\right)=\mathrm{C}\left(\mathrm{x}^2+1\right)^2\)
  2. B \(\mathrm{y}^2\left(\mathrm{y}^2+2\right)=\mathrm{C}\left(\mathrm{x}^2+1\right)\)
  3. C \(\mathrm{y}^2\left(\mathrm{y}^2+2\right)=\mathrm{C}\left(\mathrm{x}^2+1\right)^2\)
  4. D \(y^2\left(y^2+1\right)=C\left(x^2+2\right)^2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{y}^2\left(\mathrm{y}^2+2\right)=\mathrm{C}\left(\mathrm{x}^2+1\right)^2\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \left(y^3+y\right)\left(x^2+1\right) d y=\left(x y^4+2 y^2 x\right) d x \\ & \Rightarrow \int \frac{y^3+y}{y^4+2 y^2} d y=\int \frac{x}{x^2+1} d x \\ & \Rightarrow \frac{1}{4} \log _e\left(y^4+2 y^2\right)=\frac{1}{2} \log _e\left(x^2+1\right)+\log C^1 \\ & \Rightarrow \log _e\left(y^4+2 y^2\right)=2 \log _e\left(C^1\right)^2\left(x^2+1\right) \\ & \Rightarrow y^4+2 y^2=\left(c^1\right)^4\left(x^2+1\right)^2 \\ & \Rightarrow y^2\left(y^2+2\right)=C\left(x^2+1\right)^2\end{aligned}\)