The particular solution of the differential e | vedclass

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Question

(B) Given differential equation: $\frac{dy}{dx} - e^x = y e^x$ Rearranging the terms: $\frac{dy}{dx} = y e^x + e^x = (y+1) e^x$ Separating the variabl

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

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(B) Given differential equation: $\frac{dy}{dx} - e^x = y e^x$ Rearranging the terms: $\frac{dy}{dx} = y e^x + e^x = (y+1) e^x$ Separating the variables: $\int \frac{dy}{y+1} = \int e^x dx$ Integrating both sides: $\log |y+1| = e^x + C$ Given the initial condition $x = 0$ and $y = 1$: $\log |1+1| = e^0 + C$ $\log 2 = 1 + C \Rightarrow C = \log 2 - 1$ Substituting $C$ back into the general solution: $\log (y+1) = e^x + \log 2 - 1$ $\log (y+1) - \log 2 = e^x - 1$ Using the property $\log a - \log b = \log \left(\frac{a}{b}\right)$: $\log \left(\frac{y+1}{2}\right) = e^x - 1$

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

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