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  • Integrating factor of xdy/dx-y = x^4-3x is: A. x B. logx C. D. –x

Integrating factor of \frac{xdy}{dx}-y=x^4-3x is:
A. x
B. logx
C. \frac{1}{x}
D. –x

 

Answers (1)

Given differential equation
$$ \Rightarrow x \frac{d y}{d x}-y=x^{4}-3 x $$
Divide though by x
$$ \\ \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{3}-3 \\ \Rightarrow \frac{d y}{d x}+\left(-\frac{1}{x}\right) y=x^{3}-3 $$
Compare
$$ \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(-\frac{1}{x}\right) \mathrm{y}=\mathrm{x}^{3}-3 \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q} $$
We get
$$ \mathrm{P}=-\frac{1}{\mathrm{x}} \text { and } \mathrm{Q}=\mathrm{x}^{3}-3 $$
The IF integrating factor is given by el $^{\mathrm{iPdx}}$
$$ \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}} \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\log \mathrm{x}} $$

\begin{aligned} &\Rightarrow e^{\int P d x}=e^{\log x^{-1}}\\ &\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log _{\mathrm{x}}^{1}}\\ \\ &\text { Hence the IF integrating factor is } &\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\frac{1}{\mathrm{x}}\end{aligned}

Option C is correct.

Posted by

infoexpert22

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