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he particular solution of the differential equation 
(dy)/(dx)-e^(x)=ye^(x), when 
x=0 and 
y=1 is

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

Full solution

Q. The particular solution of the differential equation dydxex=yex\frac{dy}{dx}-e^{x}=ye^{x}, when x=0x=0 and y=1y=1 is
  1. Rewrite Differential Equation: First, we need to rewrite the differential equation in a form that allows us to use the method of integrating factors. The equation is already in the form of a first-order linear differential equation, dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x), where P(x)=exP(x) = -e^x and Q(x)=exQ(x) = e^x.
  2. Calculate Integrating Factor: Next, we calculate the integrating factor, μ(x)\mu(x), which is e(P(x)dx)e^{(\int P(x)\,dx)}. In this case, μ(x)=e(exdx)\mu(x) = e^{(\int -e^x\,dx)}. The integral of ex-e^x with respect to xx is ex-e^x, so μ(x)=e(ex)\mu(x) = e^{(-e^x)}.
  3. Multiply by Integrating Factor: Now, we multiply the entire differential equation by the integrating factor, μ(x)\mu(x), to get μ(x)dydx+μ(x)P(x)y=μ(x)Q(x)\mu(x) \cdot \frac{dy}{dx} + \mu(x) \cdot P(x) \cdot y = \mu(x) \cdot Q(x). Substituting the values of μ(x)\mu(x), P(x)P(x), and Q(x)Q(x), we get e(ex)dydxe(ex)exy=e(ex)exe^{(-e^x)} \cdot \frac{dy}{dx} - e^{(-e^x)} \cdot e^x \cdot y = e^{(-e^x)} \cdot e^x.
  4. Recognize Derivative Form: We notice that the left side of the equation is now the derivative of (yμ(x))(y \cdot \mu(x)) with respect to xx. Therefore, we can write ddx(ye(ex))=e(ex)ex\frac{d}{dx}(y \cdot e^{(-e^x)}) = e^{(-e^x)} \cdot e^x.
  5. Integrate Both Sides: To solve for yy, we integrate both sides of the equation with respect to xx. The integral of the left side is yeexy \cdot e^{-e^x}, and the integral of the right side is eexexdx\int e^{-e^x} \cdot e^x \, dx.
  6. Re-evaluate Integrating Factor: The integral on the right side is not a standard integral and cannot be expressed in terms of elementary functions. It seems there has been a mistake in the calculation of the integrating factor or in the setup of the equation. We need to re-evaluate the integrating factor and the setup of the equation.

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