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Question
What is the particular solution to the differential equation
(dy)/(dx)=(1+y)/(x) with the initial condition
y(e)=1?

Question $\newline$ What is the particular solution to the differential equation $\frac{d y}{d x}=\frac{1+y}{x}$ with the initial condition $y(e)=1 ?$

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Find derivatives of logarithmic functions

Full solution

Q. Question

\newline

What is the particular solution to the differential equation

\frac{d y}{d x}=\frac{1+y}{x}

with the initial condition

y(e)=1 ?

Identify type of differential equation: Step $1$ : Identify the type of differential equation. We have $\frac{dy}{dx} = \frac{1+y}{x}$ . This is a first-order linear differential equation.
Rearrange to separate variables: Step $2$ : Rearrange the equation to separate variables. $\frac{dy}{1+y} = \frac{dx}{x}$ . This step involves separating the variables $y$ and $x$ on different sides of the equation.
Integrate both sides: Step $3$ : Integrate both sides. $\newline$ Integrate left side: $\int \frac{dy}{1+y} = \ln|1+y| + C_1$ . $\newline$ Integrate right side: $\int \frac{dx}{x} = \ln|x| + C_2$ .
Combine constants and solve: Step $4$ : Combine the constants and solve for $y$ . $\ln|1+y| = \ln|x| + C$ , where $C = C_2 - C_1$ .Remove the absolute values (assuming $x$ and $1+y$ are positive): $1+y = e^{(\ln|x| + C)} = x\cdot e^C$ .
Apply initial condition: Step $5$ : Apply the initial condition to find $C$ . Given $y(e) = 1$ , plug in $x = e$ : $1 + 1 = e \cdot e^C$ . $2 = e^{(1+C)}$ . Taking natural log on both sides: $\ln(2) = 1 + C$ . $C = \ln(2) - 1$ .
Write particular solution: Step $6$ : Write the particular solution using the value of $C$ . $y = x\cdot e^{(\ln(2)-1)} - 1.$ Simplify: $y = x\cdot\left(\frac{2}{e}\right) - 1.$

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