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Match the differential equations in List I to their integrating factors in List II.
The correct match is
Options:
The correct match is
Solution:
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Verified Answer
The correct answer is:
$5 \quad 1 \quad 2 \quad 3$
(P)
$$
\begin{aligned}
& \left(x^3+1\right) \frac{d y}{d x}+x^2 y=3 x^2 \\
& \Rightarrow \frac{d y}{d x}+\frac{x^2}{x^3+1} y=\frac{3 x^2}{x^3+1} \\
& \therefore \text { Integrating factor }=e^{\int \frac{x^2}{x^3+1} d x} \\
& =e^{\frac{1}{3} \log \left(x^3+1\right)}=\left(x^3+1\right)^{1 / 3}
\end{aligned}
$$
(Q)
$$
\begin{aligned}
& x^2 \frac{d y}{d x}+3 x y=x^6 \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{3 x}{x^2} y=\frac{x^6}{x^2} \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{3}{x} y=x^4 \\
& \therefore \quad \text { Integrating factor }=e^{\int \frac{3}{x} d x} \\
& =e^{3 \log x} \\
& =x^3 \\
&
\end{aligned}
$$
(R)
$$
\begin{aligned}
& \text { R) }\left(x^3+1\right)^2 \frac{d y}{d x}+6 x^2\left(x^3+1\right) y=x^2 \\
& \begin{aligned}
& \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2\left(x^3+1\right)}{\left(x^3+1\right)^2} y=\frac{x^2}{\left(x^3+1\right)^2} \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2}{x^3+1} y=\frac{x^2}{\left(x^3+1\right)^2} \\
& \therefore \text { Integrating factor }=e^{\int \frac{6 x^2}{x^3+1}} d x \\
&=e^{2 \log \left(x^3+1\right)}=\left(x^3+1\right)^2
\end{aligned}
\end{aligned}
$$
(S)
$$
\begin{aligned}
& \left(x^2+1\right) \frac{d y}{d x}+4 x y=\ln x \\
& \Rightarrow \frac{d y}{d x}+\frac{4 x}{x^2+1} y=\frac{\ln x}{x^2+1} \\
& \therefore \text { Integrating factor }=e^{\int \frac{4 x}{x^2+1} d x} \\
& =e^{2 \log \left(x^2+1\right)} \\
& =\left(x^2+1\right)^2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \left(x^3+1\right) \frac{d y}{d x}+x^2 y=3 x^2 \\
& \Rightarrow \frac{d y}{d x}+\frac{x^2}{x^3+1} y=\frac{3 x^2}{x^3+1} \\
& \therefore \text { Integrating factor }=e^{\int \frac{x^2}{x^3+1} d x} \\
& =e^{\frac{1}{3} \log \left(x^3+1\right)}=\left(x^3+1\right)^{1 / 3}
\end{aligned}
$$
(Q)
$$
\begin{aligned}
& x^2 \frac{d y}{d x}+3 x y=x^6 \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{3 x}{x^2} y=\frac{x^6}{x^2} \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{3}{x} y=x^4 \\
& \therefore \quad \text { Integrating factor }=e^{\int \frac{3}{x} d x} \\
& =e^{3 \log x} \\
& =x^3 \\
&
\end{aligned}
$$
(R)
$$
\begin{aligned}
& \text { R) }\left(x^3+1\right)^2 \frac{d y}{d x}+6 x^2\left(x^3+1\right) y=x^2 \\
& \begin{aligned}
& \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2\left(x^3+1\right)}{\left(x^3+1\right)^2} y=\frac{x^2}{\left(x^3+1\right)^2} \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2}{x^3+1} y=\frac{x^2}{\left(x^3+1\right)^2} \\
& \therefore \text { Integrating factor }=e^{\int \frac{6 x^2}{x^3+1}} d x \\
&=e^{2 \log \left(x^3+1\right)}=\left(x^3+1\right)^2
\end{aligned}
\end{aligned}
$$
(S)
$$
\begin{aligned}
& \left(x^2+1\right) \frac{d y}{d x}+4 x y=\ln x \\
& \Rightarrow \frac{d y}{d x}+\frac{4 x}{x^2+1} y=\frac{\ln x}{x^2+1} \\
& \therefore \text { Integrating factor }=e^{\int \frac{4 x}{x^2+1} d x} \\
& =e^{2 \log \left(x^2+1\right)} \\
& =\left(x^2+1\right)^2 \\
&
\end{aligned}
$$
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