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  • Integrating factor of the differential equation 1-x^2 dy/dx -xy=1 is: A. -x

Integrating factor of the differential equation \left(1-x^{2}\right) \frac{d y}{d x}-x y=1
is:
A. -x
B. \frac{x}{1+x^{2}}
C. \sqrt{1-x^{2}}
D. \frac{1}{2} \log \left(1-x^{2}\right)

Answers (1)

\left(1-x^{2}\right) \frac{d y}{d x}-x y=1$
Divide through by \left(1-\mathrm{x}^{2}\right)$
$$ \\ \Rightarrow \frac{d y}{d x}-\frac{x y}{1-x^{2}}=\frac{1}{1-x^{2}} \\ \Rightarrow \frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}} $$
Compare
\frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}} \quad\ and \ \ \frac{d y}{d x}+P y=Q$
We get
$$ \mathrm{P}=\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}, \quad \mathrm{Q}=\frac{1}{1-\mathrm{x}^{2}} $$
The IF factor is given by
$$ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{-\mathrm{x}}{1-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}} $$
Substitute 1-x^{2}=t$ hence

\frac{d t}{d x}=-2 x$
Which means
\\-x d x=\frac{d t}{2}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{\mathrm{dt}}{2 \mathrm{t}}}$ \\$\Rightarrow e^{\int \mathrm{Pdx}}=e^{\frac{1}{2} \int \frac{\mathrm{dt}}{t}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\frac{1}{2} \log t}$ \\$\Rightarrow e^{\int P d x}=e^{\log t^{\frac{1}{2}}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\mathrm{e}^{\log \sqrt{t}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\sqrt{\mathrm{t}}$
Resubstitute
\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\sqrt{1-\mathrm{x}^{2}}$

Hence the IF integrating factor is \sqrt{1-\mathrm{x}^{2}}$

Option C is correct.

Posted by

infoexpert22

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