Question

The integrating factor of differential equation dy/dx(1+x) -xy = 1-x

Answer 1

Given,

$\displaystyle\longrightarrow\sf{(1+x)\dfrac{dy}{dx}-xy=1-x}$

Dividing each term by $\displaystyle\sf{(1+x),}$

$\displaystyle\longrightarrow\sf{\dfrac{dy}{dx}-\dfrac{x}{1+x}\cdot y=\dfrac{1-x}{1+x}}$

Comparing it with the standard linear differential equation $\displaystyle\sf{\dfrac{dy}{dx}+Py=Q,}$ we get,

$\displaystyle\longrightarrow\sf{P=-\dfrac{x}{1+x}}$

$\displaystyle\longrightarrow\sf{\int P\ dx=-\int\dfrac{x}{1+x}\ dx}$

$\displaystyle\longrightarrow\sf{\int P\ dx=-\int x\cdot\dfrac{1}{1+x}\ dx}$

$\displaystyle\longrightarrow\sf{\int P\ dx=-x\ln|1+x|+\int\ln|1+x|\ dx}$

$\displaystyle\longrightarrow\sf{\int P\ dx=-x\ln|1+x|+(1+x)\ln|1+x|-(1+x)+c}$

$\displaystyle\longrightarrow\sf{\int P\ dx=\ln|1+x|-1-x+c}$

$\displaystyle\longrightarrow\sf{\int P\ dx=\ln\left|\dfrac{1+x}{e^{1+x-c}}\right|}$

Hence the integrating factor is,

$\displaystyle\longrightarrow\sf{I.F.=e^{\int P\ dx}}$

$\displaystyle\longrightarrow\sf{I.F.=e^{\ln\left|\frac{1+x}{e^{1+x-c}}\right|}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{I.F.=\dfrac{1+x}{e^{1+x-c}}}}}$

$\displaystyle\longrightarrow\sf{I.F.=\dfrac{(1+x)e^c}{e^{1+x}}}$

Let $\displaystyle\sf{e^c=C.}$ Then,

$\displaystyle\longrightarrow\sf{\underline{\underline{I.F.=\dfrac{(1+x)C}{e^{1+x}}}}}$