Question

Which of the following is true for $y\left(x\right)$ that satisfies the differential equation $\frac{dy}{dx}=xy-1+x-y;y\left(0\right)=0$

• A. $y\left(1\right)=1$
• B. $y\left(1\right)=e^{\frac{1}{2}}-1$
• C. $y\left(1\right)=e^{\frac{1}{2}}-e^{-\frac{1}{2}}$
• D. $y\left(1\right)=e^{-\frac{1}{2}}-1$

Answer 1

The correct option is D $y\left(1\right)=e^{-\frac{1}{2}}-1$

$\frac{dy}{dx}=(x-1)y+(x-1)\Rightarrow \frac{dy}{dx}=(x-1)(y+1)\Rightarrow \frac{dy}{y+1}=(x-1)dx\Rightarrow \ln |y+1|=\frac{x^2}{2}-x+c$
$x=0,y=0$
$\Rightarrow c=0\ln |y+1|=\frac{x^2}{2}-x\Rightarrow y=-1\pm e^{\frac{x^2}{2}-x}$
But we know $x=0,y=0$, so
$y=-1+e^{\frac{x^2}{2}-x}\therefore y\left(1\right)=e^{-\frac{1}{2}}-1$​​​​​​