TS EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1\) is
- A \(y=\frac{1}{x^2}+3 c\)
- B \((3 y-1) x^3+3 y=c\)
- C \(\log y-x y=c\)
- D \(\left(1+x^3\right) y=x^3+c\)
Answer & Solution
Correct Answer
(B) \((3 y-1) x^3+3 y=c\)
Step-by-step Solution
Detailed explanation
Given differential equation is \[ \begin{array}{ll} \Rightarrow & \left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1 \\ \Rightarrow & \left(\frac{1+x^3}{x^2}\right) \frac{d y}{d x}+3 y=1 \\ \Rightarrow & \frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{x^2}{1+x^3} \end{array} \]…
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