MHT CET · Maths · Differential Equations
The integrating factor of differential equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\) is
- A \(\frac{1}{x}\)
- B \(x\)
- C \(\log x\)
- D \(e^{x}\)
Answer & Solution
Correct Answer
(B) \(x\)
Step-by-step Solution
Detailed explanation
(A)
We have \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
\(\therefore \frac{d y}{d x}=\frac{-\left(1+y+x^{2} y\right)}{x\left(1+x^{2}\right)}=\frac{-\left[1+y\left(1+x^{2}\right)\right]}{x\left(1+x^{2}\right)}\)
\(\therefore \frac{d y}{d x}=\frac{-1}{x\left(1+x^{2}\right)}-\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}\)
\(\therefore \frac{d y}{d x}+\left(\frac{1}{x}\right) y=\frac{-1}{x\left(1+x^{2}\right)}\)
\(\therefore\) I.F. \(=e^{\int \frac{1}{x} d x}=e^{\log x}=x\)
We have \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
\(\therefore \frac{d y}{d x}=\frac{-\left(1+y+x^{2} y\right)}{x\left(1+x^{2}\right)}=\frac{-\left[1+y\left(1+x^{2}\right)\right]}{x\left(1+x^{2}\right)}\)
\(\therefore \frac{d y}{d x}=\frac{-1}{x\left(1+x^{2}\right)}-\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}\)
\(\therefore \frac{d y}{d x}+\left(\frac{1}{x}\right) y=\frac{-1}{x\left(1+x^{2}\right)}\)
\(\therefore\) I.F. \(=e^{\int \frac{1}{x} d x}=e^{\log x}=x\)
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