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Integrating Factor (I.F.) of the defferential equation
\(\frac{d y}{d x}-\frac{3 x^2 y}{1+x^3}=\frac{\sin ^2(x)}{1+x} \text { is }\)
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\(\frac{d y}{d x}-\frac{3 x^2 y}{1+x^3}=\frac{\sin ^2(x)}{1+x} \text { is }\)
Solution:
1571 Upvotes
Verified Answer
The correct answer is:
\(\frac{1}{1+x^3}\)
\(\begin{aligned}
& \text {Hints: If } e^{\int p d x}=e^{-\int \frac{3 x^2 d x}{1+x^3}}=e^{-\log \left(1+x^3\right)}=e^{\log \left(1+x^3\right)^{-1}} \\
& =\left(1+x^3\right)^{-1}=\frac{1}{1+x^3}
\end{aligned}\)
& \text {Hints: If } e^{\int p d x}=e^{-\int \frac{3 x^2 d x}{1+x^3}}=e^{-\log \left(1+x^3\right)}=e^{\log \left(1+x^3\right)^{-1}} \\
& =\left(1+x^3\right)^{-1}=\frac{1}{1+x^3}
\end{aligned}\)
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