Question

The particular solution of the differential equation $\frac{dy}{dx}-e^x = ye^x$, when $x = 0$ and $y = 1$ is

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

Answer 1

Answer: (A)

$\log (\frac{y+1}{2}) = e^x - 1$

Answer 2

Given the differential equation:

$$\frac{dy}{dx} - e^x = ye^x$$

Rearrange to:

$$\frac{dy}{dx} = e^x(y+1)$$

Separate variables:

$$\frac{dy}{y+1} = e^x \, dx$$

Integrate both sides:

$$\int \frac{dy}{y+1} = \int e^x \, dx \quad \Rightarrow \quad \ln|y+1| = e^x + C$$

Apply the initial condition $x=0$ and $y=1$:

$$\ln|1+1| = e^0 + C \quad \Rightarrow \quad \ln 2 = 1 + C \quad \Rightarrow \quad C = \ln2 - 1$$

Thus, the particular solution becomes:

$$\ln|y+1| = e^x + \ln2 - 1 \quad \Rightarrow \quad \ln\left(\frac{y+1}{2}\right) = e^x - 1$$