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TS EAMCET · Maths · Differential Equations

The general solution of \(\frac{d y}{d x}=\frac{x^3\left(y^4+1\right)}{[}\) is \[ \left[2 y^{-\frac{2}{3}}+3\left(\frac{x}{\sqrt[3]{y}}\right)^2\right]^{\frac{3}{2}} \]

  1. A \(\log \left(\frac{y^4}{1+y^4}\right)=\frac{4}{9}\left(\frac{4+3 x^2}{\sqrt{2+3 x^2}}\right)+C\)
  2. B \(\frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right)=\frac{1}{9} \log \left(\frac{4+3 x^2}{\sqrt{2+3 x^2}}\right)+C\)
  3. C \(\frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right)=\frac{4}{9} \frac{1}{\sqrt{2+3 x^2}}+C\)
  4. D \(\log \left(\frac{y^4}{1+y^4}\right)=\frac{1}{9} \frac{1}{\sqrt{2+3 x^2}}+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\log \left(\frac{y^4}{1+y^4}\right)=\frac{4}{9}\left(\frac{4+3 x^2}{\sqrt{2+3 x^2}}\right)+C\)

Step-by-step Solution

Detailed explanation

Given, \(\begin{aligned} \frac{d y}{d x} & =\frac{x^3\left(y^4+1\right)}{\left[2 y^{-\frac{2}{3}}+3\left(\frac{x}{\sqrt[3]{y}}\right)^2\right]^{3 / 2}} \\ & =\frac{x^3\left(y^4+1\right)}{\left[2 y^{-2 / 3}+3 x^2 y^{-2 / 3}\right]^{3 / 2}}\end{aligned}\)…
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