Question

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

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

Answer 1

The correct option is D

$\mathrm{y}\left(1\right)={\mathrm{e}}^{-\frac{1}{2}}-1$

Step 1: Simplify the Right-hand side.

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}-1+\mathrm{x}-\mathrm{y}$

$\Rightarrow$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}-\mathrm{y}+\mathrm{x}-1$

$\Rightarrow$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left(\mathrm{x}-1\right)+1\left(\mathrm{x}-1\right)$

$\Rightarrow$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{x}-1\right)\left(\mathrm{y}+1\right)$

$\Rightarrow$$\frac{\mathrm{dy}}{\left(\mathrm{y}+1\right)}=(\mathrm{x}-1)\mathrm{dx}$

Step 2: Apply the antiderivative.

The antiderivative formulas

$\int \frac{1}{x}dx=\ln \left(x\right)+c$

$\int xdx=\frac{x^2}{2}+c$

$\int kdx=kx+c$ where $k$ is a constant

Apply antiderivative formula

$\int \frac{d\mathrm{y}}{\left(\mathrm{y}+1\right)}=\int \left(\mathrm{x}-1\right)d\mathrm{x}$

$\ln \left(\mathrm{y}+1\right)=\frac{{\mathrm{x}}^2}{2}-\mathrm{x}+\mathrm{C}$

Step 3: Put $\mathrm{x}=0$ and $\mathrm{y}=0$ in the above equation

$\ln \left(\mathrm{y}+1\right)=\frac{{\mathrm{x}}^2}{2}-\mathrm{x}+\mathrm{C}$

$\Rightarrow$$\ln \left(0+1\right)=\frac{0}{2}-0+\mathrm{C}$

$\Rightarrow$ $\mathrm{C}=0$

The equation becomes $\ln \left(\mathrm{y}+1\right)=\frac{{\mathrm{x}}^2}{2}-\mathrm{x}$

Step 4: Put $\mathrm{x}=1$ in the equation $\ln \left(\mathrm{y}+1\right)=\frac{{\mathrm{x}}^2}{2}-\mathrm{x}$

$\ln \left(\mathrm{y}+1\right)=\frac{1}{2}-1$

$\Rightarrow$$\ln \left(\mathrm{y}+1\right)=-\frac{1}{2}$

$\Rightarrow$ $\mathrm{y}+1={\mathrm{e}}^{-\frac{1}{2}}$

$\Rightarrow$ $\mathrm{y}={\mathrm{e}}^{-\frac{1}{2}}-1$

Hence, option d) $\mathrm{y}\left(1\right)={\mathrm{e}}^{-\frac{1}{2}}-1$is true.