Question

Solve y'=x-y / x, y(1)=1​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{\dfrac{dy}{dx}=\dfrac{x-y}{x}}$

$\bf{Put\,\,\,y=vx}$

$\bf{\mapsto\,\,\,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}}$

So,

$\rm{v+x\dfrac{dv}{dx}=\dfrac{x-vx}{x}}$

$\rm{\implies\,v+x\dfrac{dv}{dx}=1-v}$

$\rm{\implies\,x\dfrac{dv}{dx}=1-2v}$

$\rm{\implies\,\dfrac{dv}{2v-1}=-\dfrac{dx}{x}}$

$\rm{\implies\,\dfrac{1}{2}\cdot\dfrac{2}{2v-1}\,dv=-\dfrac{dx}{x}}$

Integrating both sides,

$\displaystyle\rm{\implies\,\dfrac{1}{2}\int\dfrac{2}{2v-1}\,dv=-\int\dfrac{dx}{x}}$

$\displaystyle\rm{\implies\,\dfrac{1}{2}\ln\left|2v-1\right|=-\ln|x|+c}$

$\displaystyle\rm{\implies\,\ln\left|2v-1\right|+2\ln|x|=2c}$

$\displaystyle\rm{\implies\,\ln\left|\dfrac{2y}{x}-1\right|+2\ln|x|=2c}$

$\displaystyle\rm{\implies\,\ln\left|\dfrac{2y-x}{x}\right|+2\ln|x|=2c}$

$\displaystyle\rm{\implies\,\ln\left|2y-x\right|-\ln|x|+2\ln|x|=2c}$

$\displaystyle\rm{\implies\,\ln\left|2y-x\right|+\ln|x|=2c}$

Since y(1) = 1, so,

$\rm{\implies\,\ln\left|2-1\right|+\ln|1|=2c}$

$\rm{\implies\,\ln\left|1\right|+\ln|1|=2c}$

$\rm{\implies\,0=2c}$

$\rm{\implies\,c=0}$

So,

$\displaystyle\rm{\implies\,\ln\left|2y-x\right|+\ln|x|=0}$

$\displaystyle\rm{\implies\,\ln\left|(2y-x)x\right|=0}$

$\displaystyle\rm{\implies\,(2y-x)x=e^{0}}$

$\displaystyle\rm{\implies\,2xy-{x}^{2}=1}$

$\displaystyle\rm{\implies\,{x}^{2}-2xy+1=0}$