KCET · Maths · Differential Equations
The integrating factor of the differential equation \( \left(2 x+3 y^{2}\right) d y=y d x(y>0) \) is
- A \( -\frac{1}{y^{2}} \)
- B \( \frac{1}{y^{2}} \)
- C \( e^{\frac{1}{y}} \)
- D \( \frac{1}{x} \)
Answer & Solution
Correct Answer
(B) \( \frac{1}{y^{2}} \)
Step-by-step Solution
Detailed explanation
\( (\mathrm{B}) \)
\( \Rightarrow \frac{\left.2 x+3 y^{2}\right) \mathrm{dy}}{y}=\mathrm{y} \mathrm{dx} \)
\( \Rightarrow \frac{2 x}{y}+3 y=\frac{d x}{d y} \)
\( \Rightarrow \frac{d x}{d y}-\frac{2 x}{y}=3 y \)
If \( =e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log y^{-2}}=\frac{1}{y^{2}} \)
\( \Rightarrow \frac{\left.2 x+3 y^{2}\right) \mathrm{dy}}{y}=\mathrm{y} \mathrm{dx} \)
\( \Rightarrow \frac{2 x}{y}+3 y=\frac{d x}{d y} \)
\( \Rightarrow \frac{d x}{d y}-\frac{2 x}{y}=3 y \)
If \( =e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log y^{-2}}=\frac{1}{y^{2}} \)
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